A non-zero vector ${{\bf{\vec a}}}$ is parallel to the line of intersection of the plane determined by vectors $\widehat {\bf{i}},\widehat {\bf{i}} - \widehat {\bf{j}}$ and the plane determined by the vectors $\widehat {\bf{i}} + \widehat {\bf{j}}$,$\widehat {\bf{i}} - {\bf{\hat k}}$. The angle between ${{\bf{\vec a}}}$ and $\widehat {\bf{i}} + 2\widehat {\bf{j}} - 2{\bf{\hat k}}$ is

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

1 of 1

Analyzing +4 | -1

A non-zero vector

{\mathbf{\vec a}}

is parallel to the line of intersection of the plane determined by vectors

\widehat \mathbf{i},\widehat \mathbf{i} - \widehat \mathbf{j}

and the plane determined by the vectors

\widehat \mathbf{i} + \widehat \mathbf{j}

\widehat \mathbf{i} - \mathbf{\hat k}

. The angle between

{\mathbf{\vec a}}

and

\widehat \mathbf{i} + 2\widehat \mathbf{j} - 2\mathbf{\hat k}

View answer and explanation

Correct option: C

\frac{\pi }{4}

Option C (

\frac{\pi }{4}

) is correct because the angle between the vector parallel to the line of intersection and the given vector is calculated using the dot product formula.

More Questions from DOT PRODUCT OF VECTORS

Practice All Chapter MCQs →

The distance between the planes given by

\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

and

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

is A vector which makes equal angles with the vectors

\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right)

, and

{\hat j}

, is If vector

\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and vector

\mathbf{b}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}

, then

\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}}

is equal to The angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

\frac{{5\pi }}{6}

and the projection of

{\mathbf{\vec a}}

in the direction of

{\mathbf{\vec b}}

\frac{{ - 6}}{{\sqrt 3 }}

then

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

is equal to A velocity

\frac{1}{4} \mathrm{~m} / \mathrm{s}

is resolved into two components along

O A

and

O B

making angles

30^{\circ}

and

45^{\circ}

respectively with the given velocity. Then the component along

O B

is Find the equation of the perpendicular drown from the origin to the plane

2x + 4y - 5z = 10