Dot product of vectors practice · Mixed difficulty

If $x+y+z=0,|\mathbf{x}|=|\mathbf{y}|=|\mathbf{z}|=2$ and $\theta$ is angle between $y$ and $\mathbf{z}$, then the value of $\operatorname{cosec}^2 \theta+\cot ^2 \theta$ is equal to

If |$\mathbf{\vec a} \times \mathbf{\vec b}| = 4$ and $\left| {\mathbf{\vec a}\cdot\mathbf{\vec b}} \right| = 2,$ then ${\left| {\mathbf{\vec a}} \right|^2}\mathrm{\;}{\left| {\mathbf{\vec b}} \right|^2}\mathrm{\;}$ is equal to

A non-zero vectors ${\mathbf{\vec a}}$ is such that its projection along the vectors $\frac{{\mathbf{\hat i} + \mathbf{\hat j}}}{{\sqrt 2 }}$ and $\frac{{ - \mathbf{\hat i} + \mathbf{\hat j}}}{{\sqrt 2 }}$ and ${\mathbf{\vec k}}$ are equal, then unit vector along ${\mathbf{\vec a}}$ is

If the vectors $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}$ are mutually orthogonal, then $(\lambda, \mu)$ is equal to

Let the unit vectors $\vec{a}$ and $\vec{b}$ be perpendicular to each other and the unit vector $\vec{c}$ be inclined at an angle ' $\theta$ 'to both $\vec{a}$ and $\vec{b}$. If $\vec{c}=\alpha \vec{a}+\beta \vec{b}+\gamma(\vec{a} \times \vec{b})$ then,

The vector equation of the plane passing through the intersection of the planes $r \cdot (\widehat i + \widehat j + \widehat k) = 1$ and $r \cdot (\widehat i - 2\widehat j) = - 2$ and the point $(1,\,\,\,0,\,\,\,2)$ is

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

The vector $\mathbf{2} \mathbf{i}+\mathbf{a} \mathbf{j}+\mathbf{k}$ is perpendicular to the vector $2 \mathbf{i}-\mathbf{j}-\mathbf{k}$, if $a=$

Let ${u}=2 {i}-{j}+3 {k}$ and ${a}=4 {i}-{j}+2 {k}$. The vector component of ${u}$ orthogonal to ${a}$ is

If $4 \mathbf{i}+\mathbf{j}-\mathbf{k}$ and $3 \mathbf{i}+m \mathbf{j}+2 \mathbf{k}$ are at right angle, then $m=$

If $\left| {\mathbf{\vec a}} \right| = \left| {\mathbf{\vec b}} \right| = 1\mathrm{\;and\;}\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right| = \sqrt 3 ,$ then the value of $\left( {3\;\mathbf{\vec a} - 4\mathbf{\vec b}} \right)\cdot\left( {2\mathbf{\vec a} + 5\mathbf{\vec b}} \right)$ is

The unit vector perpendicular to \widehat i - \widehat j$\widehat \mathbf{i} - \widehat \mathbf{j}$ and coplanar with \widehat i + 2\widehat j$\widehat \mathbf{i} + 2\widehat \mathbf{j}$ and \widehat i + 3\widehat j$\widehat \mathbf{i} + 3\widehat \mathbf{j}$ is

The unit normal vector to the line joining $\mathbf{i}-\mathbf{j}$ and $2 \mathbf{i}+3 \mathbf{j}$ and pointing towards the origin is

$(\mathbf{r} \cdot \hat{\mathbf{i}})^2+(\mathbf{r} \cdot \hat{\mathbf{j}})^2+(\mathbf{r} \cdot \hat{\mathbf{k}})^2$ is equal to

If $|{a}|=1,|{b}|=2$ and $|{a}-2 {b}|=4$ then $|{a}+3 {b} \mid$ is equal to

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\overrightarrow{{c}}=3 \hat{{i}}-\hat{{j}}+\lambda \hat{{k}}$ be three vectors. Let $\overrightarrow{{r}}$ be a unit vector along $\vec{b}+\vec{c}.$ If $\vec{r} \cdot \vec{a}=3,$ then $3 \lambda$ is equal to:

If the shortest distance between the line $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda_1(2 \hat{i}+3 \hat{j}+4 \hat{k})$ and $\vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda_2$ $(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}})$ is $x$, then $\cos ^{-1}(\cos \sqrt{6} \mathrm{x})$ is equal to

The vectors $\mathbf{A B}=3 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k}$ and $\mathbf{B C}=-\mathbf{i}-2 \mathbf{k}$ are the adjacent sides of a parallelogram. An angle between its diagonals is

Given, two vectors are $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$, the unit vector coplanar with the two vectors and perpendicular to first is

The two variable vectors 3x\widehat i + y\widehat j - 3\hat k$3x\widehat \mathbf{i} + y\widehat \mathbf{j} - 3\mathbf{\hat k}$ and x\widehat i - 4 y \widehat j + 4 \hat k$x\widehat \mathbf{i} - 4\;y\;\widehat \mathbf{j} + 4\;\mathbf{\hat k}$ are orthogonal to each other, then the locus of $\left( {x,y} \right)$ is

The equation of the plane passing through the points $(-1,-2,0),(2,3,5)$ and parallel to the line
$\mathbf{r}=-3 \mathbf{j}+\mathbf{k}+\boldsymbol{\lambda}(2 \mathbf{i}+5 \mathbf{j}-\mathbf{k})$ is

If $\mathbf{a}=2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{b}=5 \mathbf{i}-3 \mathbf{j}+\mathbf{k}$, then the projection of b on a is

If $\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c}$ are three non-zero, non-coplanar vectors and

${{\mathbf{\vec b}}_1} + \mathbf{\vec b} - \frac{{\mathbf{\vec b}\cdot\mathbf{\vec a}}}{{|\mathbf{\vec a}{|^2}}}\mathbf{\vec a}\;\;\;\;\;\;\;,{{\mathbf{\vec b}}_2} + \mathbf{\vec b} - \frac{{\mathbf{\vec b}\cdot\mathbf{\vec a}}}{{|\mathbf{\vec a}{|^2}}}\mathbf{\vec a}\;\;\;$

And

$\mathrm{\;}{{\mathbf{\vec c}}_1} + \mathbf{\vec b} - \frac{{\mathbf{\vec c}\cdot\mathbf{\vec a}}}{{|\mathbf{\vec a}{|^2}}}\mathbf{\vec a} + \frac{{\mathbf{\vec c}\cdot\mathbf{\vec b}}}{{|\mathbf{\vec b}{|^2}}}{{\mathbf{\vec b}}_1}\;$

${{\mathbf{\vec c}}_2} + \mathbf{\vec c} - \frac{{\mathbf{\vec c}\cdot\mathbf{\vec a}}}{{|\mathbf{\vec a}{|^2}}}\mathbf{\vec a} - \frac{{\mathbf{\vec c}\cdot{{\mathbf{\vec b}}_1}}}{{|{{\mathbf{\vec b}}_1}{|^2}}}{{\mathbf{\vec b}}_1}\;,\;\;\;$

${{\mathbf{\vec c}}_3} = \mathbf{\vec c} - \frac{{\mathbf{\vec c}\cdot\mathbf{\vec a}}}{{|\mathbf{\vec c}{|^2}}}\mathbf{\vec a} - \frac{{\mathbf{\vec c}\cdot{{\mathbf{\vec b}}_2}}}{{|\mathbf{\vec c}{|^2}}}{{\mathbf{\vec b}}_1},\;$

${{\mathbf{\vec c}}_4} = \mathbf{\vec c} - \frac{{\mathbf{\vec c}\cdot\mathbf{\vec a}}}{{|\mathbf{\vec c}{|^2}}}\mathbf{\vec a} - \frac{{\mathbf{\vec b}\cdot\mathbf{\vec c}}}{{|\mathbf{\vec b}{|^2}}}{{\mathbf{\vec b}}_1}$

Then, which of the following is a set of mutually orthogonal vectors?

$(\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i})+(\vec{r} \cdot \hat{j})(\vec{r} \times \hat{j})+(\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})$ is equal to

Let $\overrightarrow{{u}}, \overrightarrow{{v}}$ and $\overrightarrow{{w}}$ be vectors such that $\overrightarrow{{u}}+\overrightarrow{{v}}+\overrightarrow{{w}}=\overrightarrow{0}$.

If $|\vec{u}|=3,|\vec{v}|=4$ and $|\vec{w}|=5$, then $\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}$ is

If $\hat{a}, \hat{b}$ and $\hat{c}$ are unit vectors satisfying $\hat{a}-\sqrt{3} \hat{b}+\hat{c}=\overrightarrow{0}$, then the angle between the vectors $\hat{a}$ and $\hat{c}$ is :