The square of the distance of the point of intersection of the lines $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0$ and $\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})$ from the origin is:

[JEE Main 2026, 5 Apr (Shift 1)]

Question

The square of the distance of the point of intersection of the lines $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0$ and $\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})$ from the origin is:

[JEE Main 2026, 5 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: c) $17$. Let the point of intersection be $A$ $ \therefore A(1+\lambda a, 1-\lambda,-1) \equiv(4+2 \mu, 0,-1+\mu a) $ $ \therefore 1-\lambda=0 \Rightarrow \lambda=1 $ $ \mu a=0 \Rightarrow \mu=0 \ (\because a \neq 0) $ $ 1+\lambda a=4+2 \mu \Rightarrow 2 \mu-a=-3$ $ \Rightarrow a=3 $ $ \therefore A(4,0,-1) $ $ \therefore d^2=(\sqrt{16+1})^2=17$

The square of the distance of the point of intersection of the lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{…

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