Let the foot of perpendicular from a point $P\,\,\left( {1,2, - 1} \right)$ to the straight line $L \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{{ - 1}}$ be $N$. Let a line be drawn from $P$ parallel to the plane $x + y + 2z = 0$ which meets $L$ at point $Q$. If $\alpha $ is the acute angle between the lines $PN$ and $PQ$, then $\cos \,\alpha $ is equal to

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

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Analyzing +4 | -1

Let the foot of perpendicular from a point

P\,\,\left( {1,2, - 1} \right)

to the straight line

L \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{{ - 1}}

N

. Let a line be
drawn from

P

parallel to the plane

x + y + 2z = 0

which meets

L

at point

Q

. If

\alpha

is the acute angle between
the lines

PN

and

PQ

, then

\cos \,\alpha

is equal to

View answer and explanation

Correct option: C

\frac{1}{{\sqrt 3 }}

Option C (

\frac{1}{{\sqrt 3 }}

) is correct because it is the cosine of the acute angle between lines PN and PQ, calculated using the dot product and magnitudes of the vectors.

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\left| {\mathbf{\vec a}} \right|

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O B

30^{\circ}

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