Let the foot of perpendicular from a point \(P(1,2\), \(-1)\) to the straight line \(L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}\) be \(N\). Let a line be drawn from \(P\) parallel to the plane \(x+y+\) \(2 z=0\) which meets \(L\) at point \(Q\). If \(\alpha\) is the acute

angle between the lines \(P N\) and \(P Q\), then \(\cos \alpha\) is equal to:

[JEE Main 2021, 25 Jul (Shift 1)]

Question

Let the foot of perpendicular from a point \(P(1,2\), \(-1)\) to the straight line \(L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}\) be \(N\). Let a line be drawn from \(P\) parallel to the plane \(x+y+\) \(2 z=0\) which meets \(L\) at point \(Q\). If \(\alpha\) is the acute

angle between the lines \(P N\) and \(P Q\), then \(\cos \alpha\) is equal to:

[JEE Main 2021, 25 Jul (Shift 1)]

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Let the foot of perpendicular from a point \(P(1,2\), \(-1)\) to the straight line \(L: \frac{x}{1}=\frac{y}{0}=\frac{z}…

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