Vector Algebra

$$\begin{gathered} \text{Given,} \\ \overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}, \\ \overrightarrow{b}=\overrightarrow{a}\times \left(\hat{i}-2\hat{k}\right) \\ \text{and }\overrightarrow{c}=\overrightarrow{b}\times \hat{k} \\ \text{Now, } \\ \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}}\times \left(\hat{\mathrm{i}}-3\hat{\mathrm{k}}\right) \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& 0& -2\end{array}\right|=2\hat{i}+8\hat{j}+\hat{k} \\ \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\times \hat{\mathrm{k}}=8\hat{\mathrm{i}}-2\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{c}}-2\hat{\mathrm{j}}=8\hat{\mathrm{i}}-4\hat{\mathrm{j}} \\ \text{ Projection of }\left(\overrightarrow{\mathrm{c}}-2\hat{j}\right)\text{ on }\overrightarrow{a} \\ \left(\overrightarrow{\mathrm{c}}-2\hat{\mathrm{j}}\right)·\hat{\mathrm{a}}=\frac{⟨8,-4,0⟩·⟨3,-1,2⟩}{\sqrt{14}} \\ =\frac{28}{\sqrt{14}}=2\sqrt{14} \end{gathered}$$

Sol: Let $\overline{c} = x\overline{a} + y\overline{b}$

$$\begin{array}{l} \left| \overline{a} \right| = \sqrt{6} \\ \left| \overline{b} \right| = \sqrt{6} \\ \overline{a}.\overline{b} = 3 \\ \end{array}$$

Apply let product with $\overline{a}$

$$\begin{array}{l} \overline{a}.\overline{c} = x\overline{a}^2 + y(\overline{a}.\overline{b}) \\ 0 = 6x + 3y \Rightarrow y = -2x \quad \dots \dots 1 \\ \overline{c}^2 = x^2\overline{a}^2 + y^2\overline{b}^2 + 2xt(\overline{a}.\overline{b}) \\ 1 = 6x^2 + 6y^2 + 6xy \quad \dots \dots 2 \\ \end{array}$$

By solving 1 & 2

$$\begin{array}{l}x=\pm \frac{1}{3\sqrt{2}}y=\mp \frac{2}{3\sqrt{2}}\\ \therefore \bar{c}=\frac{1}{\sqrt{2}}(-\widehat{i}+\widehat{k})\end{array}$$

$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{b} \\ (\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=0 \\ \overrightarrow{c}=\lambda (\overrightarrow{a}+\overrightarrow{b})=\lambda (5\hat{l}-6\hat{ı}+4\hat{k}) \\ |\overrightarrow{c}{|}^2={\lambda }^2(77\left) \\ \right(\overrightarrow{a}+\overrightarrow{c})·(\overrightarrow{c}+\overrightarrow{b})=168 \\ \Rightarrow \overrightarrow{a}·\overrightarrow{c}+\overrightarrow{a}·\overrightarrow{b}+|\overrightarrow{c}{|}^2+\overrightarrow{c}·\overrightarrow{b}=168 \\ 14+\overrightarrow{c}·(\overrightarrow{a}+\overrightarrow{b})+|\overrightarrow{c}{|}^2=168 \\ 14+c|\overrightarrow{a}+{\left.\overrightarrow{b}\right|}^2+|\overrightarrow{c}{|}^2=168 \\ 14+77\lambda +77{\lambda }^2=168 \\ {\lambda }^2+\lambda -2=0 \\ \lambda =-2 or \lambda =1 \\ |\stackrel{\rightharpoonup }{c}{|}^2=77{\lambda }^2=308 or 77 \end{gathered}$$

\(\vec{r} \times \vec{a}-\vec{b} \times \vec{a}=\overrightarrow{0}\)

\((\vec{r}-\vec{b}) \times \vec{a}=\overrightarrow{0}\)

\(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}}\)

\(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{a}}\)

\( \vec{r} \cdot \vec{a}=0 \Rightarrow \vec{a} \cdot \vec{b}+\lambda|\vec{a}|^2=0\)

\(\lambda=-\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|^2}=-\frac{(1-2)}{9}=\frac{1}{9} \)

\( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\frac{\overrightarrow{\mathrm{a}}}{9} \)

\( |3 \vec{r}|^2=9|\vec{r}|^2=9\left(|\vec{b}|^2+\frac{|\vec{a}|^2}{81}+\frac{2(\vec{a} \cdot \vec{b})}{9}\right)=44\)

\(\begin{aligned}
&\begin{aligned}
& \vec{A} \cdot \vec{B}=0 \\
& (2 \hat{i}+3 n \hat{\jmath}+2 \hat{k}) \cdot(2 \hat{i}-2 \hat{\jmath}+4 p \hat{k})=0 \\
& 4-6 n+8 p=0 \\
& 8 p-6 n=-4 \\
& \quad 3 n - 4 p=2 ....\text { (1) } \\
& \text { Given equidistance from origin}
\end{aligned}\\
&\begin{gathered}
|\vec{A}|=|\vec{B}| \\
|\vec{A}|^2=|\vec{B}|^2 \\
4+9 n^2+4=4+4+16 p^2 \\
9 n^2=16 p^2 \\
3 n= \pm 4 p \\
3 n=-4 p \quad or \quad 3 n=4 p \\
8 p=-2 \\
p=-1 / 4 \\
4\left(-\frac{1}{4}\right)-3 n=-2 \\
1-3 n=-2 \\
3 n=1 \\
n=\frac{1}{3} \\
3 n+4 p=3\left(\frac{1}{3}\right)+4\left(-\frac{1}{4}\right) \\
=0
\end{gathered}
\end{aligned}\)0

$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{b} \\ (\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=0 \\ \overrightarrow{c}=\lambda (\overrightarrow{a}+\overrightarrow{b})=\lambda (5\hat{l}-6\hat{ı}+4\hat{k}) \\ |\overrightarrow{c}{|}^2={\lambda }^2(77\left) \\ \right(\overrightarrow{a}+\overrightarrow{c})·(\overrightarrow{c}+\overrightarrow{b})=168 \\ \Rightarrow \overrightarrow{a}·\overrightarrow{c}+\overrightarrow{a}·\overrightarrow{b}+|\overrightarrow{c}{|}^2+\overrightarrow{c}·\overrightarrow{b}=168 \\ 14+\overrightarrow{c}·(\overrightarrow{a}+\overrightarrow{b})+|\overrightarrow{c}{|}^2=168 \\ 14+c|\overrightarrow{a}+{\left.\overrightarrow{b}\right|}^2+|\overrightarrow{c}{|}^2=168 \\ 14+77\lambda +77{\lambda }^2=168 \\ {\lambda }^2+\lambda -2=0 \\ \lambda =-2 or \lambda =1 \\ |\stackrel{\rightharpoonup }{c}{|}^2=77{\lambda }^2=308 or 77 \end{gathered}$$

Since edges AB, AC, AD are mutually perpendicular,

we can place point $A$ at the origin $(0,0,0)$, and assign coordinates along the 3 axes:

Let $AB = x$, $AC = y$, $AD = z$

Coordinates: $A(0,0,0)$, $B(x,0,0)$, $C(0,y,0)$, $D(0,0,z)$

$\text{Area of }\mathrm{\Delta }ABC=\frac{1}{2}xy=5\text{\hspace{0.05em}⇒\hspace{0.05em}xy=10}$

$\text{Area of }\mathrm{\Delta }ACD=\frac{1}{2}yz=6\text{\hspace{0.05em}⇒\hspace{0.05em}yz=12}$

$\text{Area of }\mathrm{\Delta }ADB=\frac{1}{2}zx=7\text{\hspace{0.05em}⇒\hspace{0.05em}zx=14}$

For a tetrahedron with 3 mutually perpendicular edges from a vertex,

we know that the square of the area of the opposite face ($\Delta BCD$) is equal to the sum of squares of the areas of the three adjacent right faces.

${\left(\text{Area of }\mathrm{\Delta }BCD\right)}^2={\left(\text{Area of }\mathrm{\Delta }ABC\right)}^2+{\left(\text{Area of }\mathrm{\Delta }ACD\right)}^2+{\left(\text{Area of }\mathrm{\Delta }ADB\right)}^2$

$\left(\text{Area}\right)^2 = 5^2 + 6^2 + 7^2 = 25 + 36 + 49 = 110$

Let \(\vec a,\vec b\) be the position vectors of \(P,Q\), and \(\vec a+\vec b\) be the position vector of \(R\).

Given \(|\vec a+\vec b|=\sqrt{21}\), so \(|\vec a+\vec b|^2=21\) ...(1)

\(|\vec a-\vec b|=3\), so \(|\vec a-\vec b|^2=9\) ...(2)

Also \(\vec a\) and \(\vec a-\vec b\) are perpendicular.

So \(\vec a\cdot(\vec a-\vec b)=0\)

\(\Rightarrow |\vec a|^2-\vec a\cdot\vec b=0\)

\(\Rightarrow \vec a\cdot\vec b=|\vec a|^2\) ...(3)

from (1), (2) and (3):

\(\vec a\cdot\vec b=|\vec a|^2=3\)

and \(|\vec b|^2=12\)

Area of triangle \(OPR\) is

\(\dfrac12|\vec a\times(\vec a+\vec b)|\)

\(=\dfrac12|\vec a\times\vec b|\)

Now \(|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2\)

\(=3\cdot12-3^2=36-9=27\)

\(|\vec a\times\vec b|=3\sqrt3\)

Area \(=\dfrac12\cdot3\sqrt3=\dfrac{3\sqrt3}{2}\)

Given: $\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k},\overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k},\overrightarrow{c}=\lambda \widehat{i}+\widehat{j}+\widehat{k}$

And, $\overrightarrow{v}=\overrightarrow{a}\times \overrightarrow{b}$

Then, \(\vec{v}=(\vec{a} \times \vec{b})=(-\hat{i}+7 \hat{j}+5 \hat{k})\)

Now, \(\vec{v} \cdot \vec{c}=11\)

\(\Rightarrow (-\hat{i}+7 \hat{j}+5 \hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\hat{k})=11\)

$\Rightarrow -\lambda +7+5=11$

$\Rightarrow \lambda =1$

Length of projection of $\overrightarrow{b}$ on $\overrightarrow{c}=\overrightarrow{b}\cdot \widehat{c}$

\(\Rightarrow\left|(2 \hat{i}+\hat{j}-\hat{k}) \cdot \frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\right|=\frac{2+1-1}{\sqrt{3}}\)

\(p=\frac{2}{\sqrt{3}}\)

$\Rightarrow 9p^2=9\left(\frac{4}{3}\right)=12$

Given \(\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=\hat{i}-\hat{j}-3 \hat{k} \)

\(a^2=\vec{a} \cdot \vec{a}=1+1+4=6\)

\(b^2=\vec{b} \cdot \vec{b}=1+1+9=11\)

\(\vec{a} \cdot \vec{b}=-1-1-6=-8\)

\(\vec{d}=\vec{c} \times \vec{a}\)

\( \vec{d}=(\vec{a} \times \vec{b}) \times \vec{a} \)
\( \vec{d}=(a^2) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{a} \)
\( \vec{d}=6 \vec{b}+8 \vec{a} \)
\( (\vec{a}-\vec{b}) \cdot \vec{d}=(\vec{a}-\vec{b}) \cdot(6 \vec{b}+8 \vec{a}) \)
\( =8 a^2-6 b^2-2 \vec{a} \cdot \vec{b} \)
\(=48-66+16=-2\)

Since \(\vec{X}, \vec{Y}, \vec{Z}\) lie in a plane, their scalar triple product is \(0\). Hence the determinant formed by these three vectors must be \(0\).

Write \(\vec{X}, \vec{Y}, \vec{Z}\) in terms of \(\vec{x}, \vec{y}, \vec{z}\):

\(\vec{X}=\alpha\vec{x}+\beta\vec{y}-\vec{z},\quad \vec{Y}=-\vec{x}+\alpha\vec{y}+\beta\vec{z},\quad \vec{Z}=\beta\vec{x}-\vec{y}+\alpha\vec{z}\).

If \(B=[\vec{x}\ \vec{y}\ \vec{z}]\) and \(A=\begin{pmatrix}\alpha&-1&\beta\\ \beta&\alpha&-1\\ -1&\beta&\alpha\end{pmatrix}\), then \([\vec{X}\ \vec{Y}\ \vec{Z}]=BA\).

Therefore, \(\det[\vec{X}\ \vec{Y}\ \vec{Z}]=\det(B)\det(A)\).

Now,

\(\det(B)=\begin{vmatrix}1&2&3\\2&3&1\\3&1&2\end{vmatrix}=1(6-1)-2(4-3)+3(2-9)=-18\neq 0\).

So, for \(\vec{X}, \vec{Y}, \vec{Z}\) to be coplanar, we must have \(\det(A)=0\).

Now,

\(\det(A)=\begin{vmatrix}\alpha&-1&\beta\\ \beta&\alpha&-1\\ -1&\beta&\alpha\end{vmatrix}\)

\(=\alpha(\alpha^2+\beta)+(\alpha\beta-1)+\beta(\beta^2+\alpha)\)

\(=\alpha^3+\beta^3+3\alpha\beta-1\).

Hence,

\(\alpha^3+\beta^3+3\alpha\beta-1=0\).

Write it as

\(\alpha^3+\beta^3+(-1)^3-3\alpha\beta(-1)=0\).

Using \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), with \(a=\alpha,\ b=\beta,\ c=-1\), we get

\((\alpha+\beta-1)(\alpha^2+\beta^2+1-\alpha\beta+\alpha+\beta)=0\).

Also,

\(\alpha^2+\beta^2+1-\alpha\beta+\alpha+\beta=\frac{1}{2}(\alpha-\beta)^2+\frac{1}{2}(\alpha+1)^2+\frac{1}{2}(\beta+1)^2>0\).

So the second factor cannot be \(0\). Therefore,

\(\alpha+\beta-1=0\Rightarrow \alpha+\beta=1\).

Hence,

\(\alpha+\beta-3=1-3=-2\).

Therefore, the correct answer is \(-2\).

Equation of anlge bisector will be

$$\begin{gathered} x-y = 0 \\ \left|\frac{a-(1-a)}{\sqrt{2}}\right|=\frac{9}{\sqrt{2}} \\ 2a-1=\pm 9 \\ a = 5, -4 \end{gathered}$$

$$\begin{gathered} \text{Given, } \overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k} \& \overrightarrow{b}=3\hat{i}+\hat{j}-\hat{k} \\ \overrightarrow{a}·\overrightarrow{b}=2 \& \overrightarrow{a}·\overrightarrow{c}=5 \& \overrightarrow{b}·\overrightarrow{c}=0 \\ \left|\overrightarrow{a}\right|=\sqrt{14}\text{ and }\left|\overrightarrow{b}\right|=\sqrt{11} \\ \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b} \\ \overrightarrow{a}·\overrightarrow{c}=\lambda |\overrightarrow{a}{|}^2+\mu \overrightarrow{a}·\overrightarrow{b} \\ 5=\lambda \times 14+\mu \times 2 ........\text{(i)} \\ \overrightarrow{b}·\overrightarrow{c}=\lambda \overrightarrow{a}·\overrightarrow{b}+\mu (\overrightarrow{b}{)}^2 \\ 0=2\lambda +11\mu ........\text{.(ii)} \\ \text{On solving, (i) \& (ii), we get} \\ \mu =-\frac{1}{15} \lambda =\frac{11}{30} \\ \text{Now}, \overrightarrow{c}=\frac{1}{6}\hat{i}+\frac{2}{3}\hat{j}+\frac{7}{6}\hat{k} \\ \left|\overrightarrow{c}\right|=\sqrt{\frac{11}{6}} \end{gathered}$$

\(\begin{equation}
\begin{aligned}
& \vec{c}=(\vec{b} \cdot \hat{a}) \hat{a}=\frac{2 \lambda-4}{9} \vec{a} \\
& \because|\vec{a}+\vec{c}|=7 \Rightarrow\left|\vec{a}\left(1+\frac{2 \lambda-4}{9}\right)\right|=7 \\
& \because \lambda>0 \Rightarrow \lambda=8 \\
& \Rightarrow \vec{c}=\frac{4}{3} \vec{a} \text { and } \vec{b}=4(2 \hat{i}+\hat{k}) \\
& \left.\Rightarrow \vec{b} \times \vec{c}=\frac{16}{3}\left|\begin{array}{ll}
\hat{i} & \vec{j} & \vec{k} \\
2 & 0 & 1\\
2 & 2 & -1
\end{array}\right|=\frac{16}{3}(-2 \hat{i}+4 \hat{j}+4 \hat{k}) \right\rvert\, \\
& \Rightarrow|\vec{b} \times \vec{c}|=\frac{32}{3}|-\hat{i}+2 \hat{j}+2 \hat{k}|=32 \\
& \Rightarrow A r e a \text { of parallelogram formed by } \vec{b} \text { and } \vec{c} \\
& \Rightarrow|\vec{b} \times \vec{c}|=32
\end{aligned}
\end{equation}\)

Given: \( \vec{a} \times \vec{b}-2(\vec{a} \times \vec{c})=0 \)

\(\Rightarrow \vec{a} \times(\vec{b}-2 \vec{c})=0\)

\(\Rightarrow \vec{b}-2 \vec{c}=\lambda \vec{a} \ldots . .(i) \)

\(\Rightarrow |\lambda \vec{a}|^2=|\vec{b}-2 \vec{c}|^2\)

\(\Rightarrow \lambda^2|\vec{a}|^2=|\vec{b}|^2+4 |\vec{c}|^2-4 \vec{b} \cdot \vec{c} \)

\(\Rightarrow \lambda^2=16+16-4.4 \cdot 2 \cdot \frac{1}{2} \)

\(\Rightarrow \lambda^2=16 \)

\(\Rightarrow \lambda= \pm 4 \)

\( \because \vec{b}-2 \vec{c}= \pm 4 \vec{a}\)

Dot with \(\vec{c} \)

\(\vec{b} \cdot \vec{c}-2|\vec{c}|^2= \pm 4(\vec{a} \cdot \vec{c})\)

\(\Rightarrow 4 \cdot 2 \cdot \frac{1}{2}-2 \cdot 4= \pm 4(\vec{a} \cdot \vec{c}) \)

\(\Rightarrow |\vec{a} \cdot \vec{c}|=1\)

Equation of anlge bisector will be

$$\begin{gathered} x-y = 0 \\ \left|\frac{a-(1-a)}{\sqrt{2}}\right|=\frac{9}{\sqrt{2}} \\ 2a-1=\pm 9 \\ a = 5, -4 \end{gathered}$$

\(2(\vec{a} \times \vec{b})+3(\vec{b} \times \vec{c})=\overrightarrow{0}\)

\((2 \vec{a}-3 \vec{c}) \times \vec{b}=\overrightarrow{0}\)

\(\vec{b} \|(2 \vec{a}-3 \vec{c})\)

\(2 \vec{a}-3 \vec{c}=\lambda \vec{b}\)

\(\overrightarrow{\mathrm{c}}=\frac{2 \overrightarrow{\mathrm{a}}-\lambda \overrightarrow{\mathrm{b}}}{3}\)

\(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=15\)

\(\left(\frac{2 \vec{a}-\lambda \vec{b}}{3}\right) \cdot \vec{a}=15\)

\(2(26)-\lambda(40-2-3)=45\)

\(\lambda=\frac{1}{5}\)

\(\Rightarrow \overrightarrow{\mathrm{c}}.(\hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})=\frac{\left(2(4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-\frac{1}{5}(10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k})\right) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})}{3}\)

\(=\frac{2(4-1-9)-\frac{1}{5}(10+2+3)}{3}=-5\)

$$\begin{gathered} \text{Given, the position vector of triangle are} \\ 4\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}-3\overrightarrow{\mathrm{r}},-5\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+2\overrightarrow{\mathrm{r}} \\ \& 2\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+2\overrightarrow{\mathrm{r}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ OG : GC=2:1 \\ \mathrm{O}\text{ (orthocentre) =}\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4} \\ \mathrm{C}\text{ (circum centre)= }\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}} \\ G\left(\mathrm{centroid}\right)=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3} \\ \text{By relation, } \\ \Rightarrow 2\left(\mathrm{\alpha }\overrightarrow{\mathrm{p}}+\mathrm{\beta }\overrightarrow{\mathrm{q}}+\mathrm{\gamma }\overrightarrow{\mathrm{r}}\right)+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right) \\ \Rightarrow 8\left(\mathrm{\alpha }\overrightarrow{\mathrm{p}}+\mathrm{\beta }\overrightarrow{\mathrm{q}}+\mathrm{\gamma }\overrightarrow{\mathrm{r}}\right)=3\left(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}\right) \\ \text{On comparing the coefficient, we get} \\ \Rightarrow 8\mathrm{\alpha }=3,8\mathrm{\beta }=3,8\mathrm{\gamma }=3 \\ \mathrm{\alpha }=\frac{3}{8},\mathrm{\beta }=\frac{3}{8},\mathrm{\gamma }=\frac{3}{8} \\ \therefore \mathrm{\alpha }+2\mathrm{\beta }+5\mathrm{\gamma }=\frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3 \end{gathered}$$

Given, vectors are

$\mathrm{a}=\mathrm{i}-\mathrm{k}, \mathrm{b}=x\mathrm{i}+\mathrm{j}+(1-x)\mathrm{k}$ and $\mathrm{c}=y\mathrm{i}+x\mathrm{j}+(1+x-y)\mathrm{k}$

$\therefore [a b c]=\left|\begin{array}{ccc}1& 0& -1\\ x& 1& 1-x\\ y& x& 1+x-y\end{array}\right|$

$\text{ Applying }{C}_3\to C_3+C_1\text{, we get }$

$$\begin{gathered} =\left|\begin{array}{ccc}1& 0& 0\\ x& 1& 1\\ y& x& 1+x\end{array}\right| \\ =1(1+x)-x \\ =1 \end{gathered}$$

$\text{ Thus, }[\mathrm{a}\text{ b c] depends upon neither }x \text{ nor }y\text{. }$

Let \(\bar{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\)

\(\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=1\)

Let \(\bar{b}=2 \vec{i}-\hat{j}+2 \hat{k}, \bar{c}=\vec{i}-2 \hat{j}-2 \hat{k}\)

\( \bar{d}=\hat{k} \)
\( \frac{\bar{a} \cdot \bar{b}}{|b|}=\frac{\bar{a} \cdot \bar{c}}{|c|}=\frac{\bar{a} \cdot \bar{d}}{|d|} \)
\( \frac{2 a_1-a_2+2 a_3}{3}=\frac{a_1+2 a_2-2 a_3}{3}=a_3\)

By solving

\(a_1=\frac{7}{\sqrt{155}}, a_2=\frac{9}{\sqrt{155}}, a_3=\frac{5}{\sqrt{155}}\)

unit vector along \(\vec{a}\):

\(\widehat{a}=\frac{1}{\sqrt{7^2+9^2+5^2}}(7 \hat{i}+9 \hat{j}+5 \hat{k})=\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})\)

$\overrightarrow{v}\cdot \overrightarrow{v}=\left|u\right|\cdot \left|v\right|\cdot \mathrm{cos\theta }$

$\Rightarrow 6-1=\sqrt{10}\cdot \sqrt{5+{\mathrm{\lambda }}^2}\cdot \frac{\sqrt{5}}{2\sqrt{7}}$

$\Rightarrow 1=\sqrt{2}\cdot \sqrt{5+{\mathrm{\lambda }}^2}\cdot \frac{1}{2\sqrt{7}}$

$\Rightarrow 14=5+{\mathrm{\lambda }}^2$

$\Rightarrow {\mathrm{\lambda }}^2=9$

$\Rightarrow \mathrm{\lambda }=3$

$v_1=k\overrightarrow{u}$

$\overrightarrow{v}={\overrightarrow{v}}_1+{\overrightarrow{v}}_2$

$\Rightarrow \overrightarrow{v}=k\overrightarrow{u}+{\overrightarrow{v}}_2$

$\overrightarrow{v}\cdot \overrightarrow{u}=k\cdot {\left|\overrightarrow{u}\right|}^2$

$\Rightarrow 5=k\cdot 10\Rightarrow k=\frac{1}{2}$

$\therefore {\overrightarrow{v}}_1=\frac{u}{2}=\frac{3\widehat{i}}{2}-\frac{\widehat{j}}{2}$

${\left|{\overrightarrow{v}}_1\right|}^2=\frac{10}{4}$

${\overrightarrow{v}}_2=\overrightarrow{v}-{\overrightarrow{v}}_1$

$=\frac{1}{2}\widehat{i}+\frac{3\widehat{j}}{2}-3\widehat{k}$

${\left|{\overrightarrow{v}}_2\right|}^2=\frac{10}{4}+9$

${\left|{\overrightarrow{v}}_1\right|}^2+{\left|{\overrightarrow{v}}_2\right|}^2=\frac{10}{4}+\frac{10}{4}+9=14$

\(\begin{aligned} \vec{a} & =\hat{i}+2 \hat{j}+\hat{k} \\ \vec{b} & =2 \hat{i}+\hat{j}-\hat{k}\end{aligned}\)

A vector \(\vec{c}\) in the plane of \(\vec{a}\) and \(b\) can be expressed as a linear combination of \(\vec{a}\) and \(\vec{b}\).

\(\vec{c}=\lambda \vec{a}+\mu \vec{b}\)

\(\vec{c} \perp \vec{a}\)

\(\vec{c} \cdot \vec{a}=0\)

\((\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{a}}=0\)

since \(\vec{a} \cdot \vec{a}=1^2+2^2+1^2=6\)

\(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}=3\)

\(\begin{aligned} & 6 \lambda+3 \mu=0 \\ & \mu=-2 \lambda\end{aligned}\)

Let \(\lambda=1\) then \(\mu=-2\)

Now \(\overrightarrow{\mathrm{c}}=3(-\hat{\mathrm{i}}+\hat{\mathrm{k}})\)

\(\hat{c}=\frac{-\hat{\mathrm{i}}+\hat{\mathrm{k}}}{\sqrt{2}}\)

Given $\overrightarrow{\mathrm{a}}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}} \text{and} \overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}-5\hat{\mathrm{j}}+3\hat{\mathrm{k}}$

$$\begin{gathered} \text{Also,} \overrightarrow{ \mathrm{a}}\times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{b}} \\ (\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=\overrightarrow{0} \\ \Rightarrow \overrightarrow{\mathrm{c}}=\lambda (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ \overrightarrow{\mathrm{c}}=\lambda (5\hat{\mathrm{i}}-6\hat{\mathrm{j}}+4\hat{\mathrm{k}})\dots .\left(i\right) \\ |\overrightarrow{\mathrm{c}}{|}^2=77{\lambda }^2 \end{gathered}$$

$$\begin{gathered} \text{Next we have, }(\overrightarrow{a}+\overrightarrow{c})·(\overrightarrow{b}+\overrightarrow{c})=168 \\ \overrightarrow{\mathrm{a}}·\overrightarrow{ \mathrm{b}}+\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}·\overrightarrow{ \mathrm{b}}+|\overrightarrow{\mathrm{c}}{|}^2=168 \\ 14+\overrightarrow{c}·(\overrightarrow{a}+\overrightarrow{b})+77{\lambda }^2=168 \\ \text{On substituting value of} \overrightarrow{c} \text{from eq} \left(i\right) \\ \lambda |5\hat{i}-6\hat{\mathrm{j}}+4\hat{\mathrm{k}}{|}^2+77{\lambda }^2=154 \\ 77\lambda +77{\lambda }^2-154=0 \\ {\lambda }^2+\lambda -2=0 \\ \lambda =-2,1 \\ \text{for} \text{ maximum value of }|\overrightarrow{\mathrm{c}}{|}^2, \text{take }\lambda =-2, \end{gathered}$$

$|\overrightarrow{\mathrm{c}}{|}^2=77{\lambda }^2=77\times 4=308$

Let

$$\vec a = x\hat i + y\hat j + z\hat k$$

and the three given vectors be

${\vec{b}}_1=2\widehat{i}-\widehat{j}+2\widehat{k},{\vec{b}}_2=\widehat{i}+2\widehat{j}-2\widehat{k},$ ${\vec{b}}_3=\widehat{k}$

Scalar projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$.

Given that the projections are equal:

$$\frac{\vec a\cdot\vec b_1}{|\vec b_1|} = \frac{\vec a\cdot\vec b_2}{|\vec b_2|} = \frac{\vec a\cdot\vec b_3}{|\vec b_3|}$$

Now $|\vec b_1| = |\vec b_2| = 3,\; |\vec b_3|=1$.

Compute dot products:

$\vec{a}\cdot {\vec{b}}_1=2x-y+2z,\vec{a}\cdot {\vec{b}}_2=x+2y-2z,$ $\vec{a}\cdot {\vec{b}}_3=z$

Conditions:

$$\frac{2x - y + 2z}{3} = \frac{x + 2y - 2z}{3}$$ $$\frac{2x - y + 2z}{3} = z$$

From the first:

$$2x - y + 2z = x + 2y - 2z \Rightarrow x - 3y + 4z = 0 \quad (1)$$

From the second:

$$2x - y + 2z = 3z \Rightarrow 2x - y - z = 0 \quad (2)$$

Solve (1) and (2):

From (2): $z = 2x - y$.

Substitute into (1):

$x-3y+4(2x-y)=0\Rightarrow x-3y+8x-4y=0$$\Rightarrow 9x-7y=0\Rightarrow x=\frac{7}{9}y$

Then $z=2x-y=2\cdot \frac{7}{9}y-y$ $=\frac{14-9}{9}y=\frac{5}{9}y$.

So

$$x : y : z = \frac{7}{9}y : y : \frac{5}{9}y = 7 : 9 : 5$$

Thus $\vec a$ is along $(7,9,5)$.

Unit vector along $\vec a$:

$\widehat{a}=\frac{1}{\sqrt{7^2+9^2+5^2}}(7\widehat{i}+9\widehat{j}+5\widehat{k})$ $=\frac{1}{\sqrt{155}}(7\widehat{i}+9\widehat{j}+5\widehat{k})$

\(\begin{aligned}
& \vec{c}=(\vec{a} \times(\hat{i}-2 \hat{j})) \times \widehat{k} \\
& =(\vec{a} \cdot \widehat{k}) (\hat{i}-2 \hat{j})-0 \\
& =-\hat{i}+2 \hat{j} \\
& \vec{c}-2 \hat{j}=-\hat{i}
\end{aligned}\)

\(\begin{aligned}
& \operatorname{Proj}=\frac{-\hat{i} \cdot \vec{a}}{|\vec{a}|} \\
& =\frac{-3}{\sqrt{14}}
\end{aligned}\)

Magnitude of the projection =$=\frac{3}{\sqrt{14}}=\frac{3\sqrt{14}}{14}$

Given, vectors are

$\mathrm{a}=\mathrm{i}-\mathrm{k}, \mathrm{b}=x\mathrm{i}+\mathrm{j}+(1-x)\mathrm{k}$ and $\mathrm{c}=y\mathrm{i}+x\mathrm{j}+(1+x-y)\mathrm{k}$

$\therefore [a b c]=\left|\begin{array}{ccc}1& 0& -1\\ x& 1& 1-x\\ y& x& 1+x-y\end{array}\right|$

$\text{ Applying }{C}_3\to C_3+C_1\text{, we get }$

$$\begin{gathered} =\left|\begin{array}{ccc}1& 0& 0\\ x& 1& 1\\ y& x& 1+x\end{array}\right| \\ =1(1+x)-x \\ =1 \end{gathered}$$

$\text{ Thus, }[\mathrm{a}\text{ b c] depends upon neither }x \text{ nor }y\text{. }$

$$\begin{gathered} \text{Given,} \\ \hat{a}·\hat{b}=\frac{1}{2} \\ Now\left(\lambda \hat{a}+2\hat{b}\right)·\left(3\hat{a}-\lambda \hat{b}\right)=0 \\ 3\lambda \hat{a}.\hat{a}-{\lambda }^2\hat{a}·\hat{b}+6\hat{a}·\hat{b}-2\lambda \hat{b}·\hat{b}=0 \\ 3\lambda -\frac{{\lambda }^2}{2}+3-2\lambda =0 \\ {\lambda }^2-2\lambda -6=0 \\ \lambda =1\pm \sqrt{7} \end{gathered}$$

Both values of \(\lambda\) does not lie in the interval \([-1,3]\)

hence number of values of \(\lambda = 0\)

Given vectors: $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$

$\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$

We are given two conditions for a vector $\vec{c}$:

1. $(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$
2. $\vec{a} \cdot \vec{c} = 3$

We need to find the value of $|\vec{a} \cdot \vec{d}|$ where $\vec{d} = \vec{b} \times \vec{c}$.

Let's analyze the first condition: $(\vec{a} - \vec{c}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{b}$

We know that the cross product is anti-commutative, i.e., $\vec{c} \times \vec{b} = -(\vec{b} \times \vec{c})$.

So, $\vec{a} \times \vec{b} - (-\vec{b} \times \vec{c}) = \vec{a} \times \vec{b} + \vec{b} \times \vec{c}$.

From the problem statement, we have $\vec{d} = \vec{b} \times \vec{c}$.

Substituting this into the expanded first condition:

$\vec{a} \times \vec{b} + \vec{d} = -18\hat{i} - 3\hat{j} + 12\hat{k}$

Now, let's calculate $\vec{a} \times \vec{b}$: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix}$

$= \hat{i}((-3)(5) - (1)(2)) - \hat{j}((2)(5) - (1)(3)) + \hat{k}((2)(2) - (-3)(3))$

$= \hat{i}(-15 - 2) - \hat{j}(10 - 3) + \hat{k}(4 + 9)$

$= -17\hat{i} - 7\hat{j} + 13\hat{k}$

Substitute this result back into the equation for $\vec{d}$:

$-17\hat{i} - 7\hat{j} + 13\hat{k} + \vec{d} = -18\hat{i} - 3\hat{j} + 12\hat{k}$

Now, solve for $\vec{d}$: $\vec{d} = (-18\hat{i} - 3\hat{j} + 12\hat{k}) - (-17\hat{i} - 7\hat{j} + 13\hat{k})$

$\vec{d} = (-18 - (-17))\hat{i} + (-3 - (-7))\hat{j} + (12 - 13)\hat{k}$

$\vec{d} = (-18 + 17)\hat{i} + (-3 + 7)\hat{j} + (12 - 13)\hat{k}$

$\vec{d} = -\hat{i} + 4\hat{j} - \hat{k}$

Finally, we need to find $|\vec{a} \cdot \vec{d}|$. We have $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{d} = -\hat{i} + 4\hat{j} - \hat{k}$.

Calculate the dot product $\vec{a} \cdot \vec{d}$: $\vec{a} \cdot \vec{d} = (2)( -1) + (-3)(4) + (1)(-1)$ $= -2 - 12 - 1$ $= -15$

The question asks for the absolute value of this dot product: $|\vec{a} \cdot \vec{d}| = |-15| = 15$.

Note that the second condition $\vec{a} \cdot \vec{c} = 3$ was not required to find $\vec{a} \cdot \vec{d}$. This condition, along with the definition of $\vec{d}$, would allow us to uniquely determine the components of $\vec{c}$ if needed.

The final answer is $\boxed{\text{15}}$.

\(|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9\)

\(|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2+|\vec{c}|^2-2 \vec{b} \cdot \vec{c}+|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=9\)

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}-=-\frac{3}{2}$

\(|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c})=0\)

$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\Rightarrow \overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}$

\(\begin{aligned} & |2 \vec{a}+k(\vec{b}+\vec{c})|=3 \\ & |\vec{a}(2-k)|=3\end{aligned}\)

\(2-k= \pm 3\)

$k=5\text{ or }-1$

Positive value of $k$ is $5$

$$\begin{gathered} \stackrel{\rightharpoonup }{a}=\hat{i}+2\hat{j}+3\hat{k} \& \overrightarrow{b}=3\hat{i}+\hat{j}-\hat{k} \\ \overrightarrow{a}.\overrightarrow{b}=2 \& \overrightarrow{a}.\overrightarrow{c}=5 \& \overrightarrow{b}.\overrightarrow{c}=0 \end{gathered}$$

\( |a|=\sqrt{14}\) and \(|b|=\sqrt{11} \quad b . c=0 \)
\( \bar{c}=\lambda \bar{a}+\mu \bar{b}\)
\( \bar{a} \cdot \bar{c}=\lambda |\bar{a}|^2+\mu \bar{a} \cdot \bar{b} \)
\( 5=\lambda 14+\mu 2 \)
\( \bar{b} \cdot \bar{c}=\lambda \bar{a} \cdot \bar{b}+\mu(b)^2 \)
\( 0=2 \lambda+11 \mu \)

\( \mu=\frac{1}{-15} \quad \lambda=\frac{11}{30}\)

$\overrightarrow{c}=\frac{1}{6}\hat{i}+\frac{2}{3}\hat{j}+\frac{7}{6}\hat{k}$

\(|c| = \sqrt{\frac{11}{6}}\)

$$\begin{gathered} \text{Given,} \\ \hat{a}·\hat{b}=\frac{1}{2} \\ Now(\lambda \hat{a}+2\hat{b})·(3\hat{a}-\lambda \hat{b})=0 \\ 3\lambda \hat{a}.\hat{a}-{\lambda }^2\hat{a}·\hat{b}+6\hat{a}·\hat{b}-2\lambda \hat{b}·\hat{b}=0 \\ 3\lambda -\frac{{\lambda }^2}{2}+3-2\lambda =0 \\ {\lambda }^2-2\lambda -6=0 \\ \lambda =1\pm \sqrt{7} \end{gathered}$$

Both values of \(\lambda\) does not lie in the interval \([-1,3]\)

hence number of values of \(\lambda = 0\)

$$\begin{gathered} \text{Given, the position vector of triangle are} \\ 4\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}-3\overrightarrow{\mathrm{r}},-5\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+2\overrightarrow{\mathrm{r}} \\ \& 2\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+2\overrightarrow{\mathrm{r}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ OG : GC=2:1 \\ \mathrm{O}\text{ (orthocentre) =}\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4} \\ \mathrm{C}\text{ (circum centre)= }\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}} \\ G\left(\mathrm{centroid}\right)=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3} \\ \text{By relation, } \\ \Rightarrow 2\left(\mathrm{\alpha }\overrightarrow{\mathrm{p}}+\mathrm{\beta }\overrightarrow{\mathrm{q}}+\mathrm{\gamma }\overrightarrow{\mathrm{r}}\right)+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right) \\ \Rightarrow 8\left(\mathrm{\alpha }\overrightarrow{\mathrm{p}}+\mathrm{\beta }\overrightarrow{\mathrm{q}}+\mathrm{\gamma }\overrightarrow{\mathrm{r}}\right)=3\left(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}\right) \\ \text{On comparing the coefficient, we get} \\ \Rightarrow 8\mathrm{\alpha }=3,8\mathrm{\beta }=3,8\mathrm{\gamma }=3 \\ \mathrm{\alpha }=\frac{3}{8},\mathrm{\beta }=\frac{3}{8},\mathrm{\gamma }=\frac{3}{8} \\ \therefore \mathrm{\alpha }+2\mathrm{\beta }+5\mathrm{\gamma }=\frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3 \end{gathered}$$

$$\begin{gathered} \text{Given,} \\ \hat{a}·\hat{b}=\frac{1}{2} \\ Now(\lambda \hat{a}+2\hat{b})·(3\hat{a}-\lambda \hat{b})=0 \\ 3\lambda \hat{a}.\hat{a}-{\lambda }^2\hat{a}·\hat{b}+6\hat{a}·\hat{b}-2\lambda \hat{b}·\hat{b}=0 \\ 3\lambda -\frac{{\lambda }^2}{2}+3-2\lambda =0 \\ {\lambda }^2-2\lambda -6=0 \\ \lambda =1\pm \sqrt{7} \end{gathered}$$