Dot product of vectors practice · Easy

In a tetrahedron $O A B C$, the edges are of lengths, $|O A|=|B C|=a,|O B|=|A C|=b,|O C|=|A B|=c$. Let $G_1$ and $G_2$ be the centroids of the triangle $A B C$ and $A O C$ such that $O G_1 \perp B G_2$, then the value of $\frac{a^2+c^2}{b^2}$ is

The angle between the straight lines \vec r = ( 2 - 3t )\widehat i + ( 1 + 2t )\widehat j + ( 2 + 6t )\hat k$\mathbf{\vec r} = \left( {2 - 3t} \right)\widehat \mathbf{i} + \left( {1 + 2t} \right)\widehat \mathbf{j} + \left( {2 + 6t} \right)\mathbf{\hat k}$ and \vec r = ( 1 + 4s )\widehat i + ( 2 - s )\widehat j + ( 8s - 1 )\hat k$\mathbf{\vec r} = \left( {1 + 4s} \right)\widehat \mathbf{i} + \left( {2 - s} \right)\widehat \mathbf{j} + \left( {8s - 1} \right)\mathbf{\hat k}$ is

If a, b and c are three mutually perpendicular vectors, then the projection of the vectors $l\frac{a}{{|a|}} + m\frac{b}{{|b|}} + n\frac{{(a \times b)}}{{|a \times b|}}$ along the angle bisector of the vectors a and b is:

If $\vec a,\overrightarrow {\;\,b} ,\;\,\vec c$ are three vectors such that $\vec a + \vec b + \vec c = 0$ and $\left| {\,\vec a\,} \right| = 2,\,\;\left| {\,\vec b\,} \right| = 3,\;\,\left| {\,\vec c\,} \right| = 4$, then the value of $\vec a \cdot \vec b\,\, + \,\,\vec b \cdot \vec c\,\, + \,\,\vec c \cdot \vec a$ is equal to

Given the vectors ${a}(2,1,-1)$, ${b}(1,2,1)$, ${c}(2,-1,3)$, ${d}(3,-1,2)$, the projection of the vector ${a}+{c}$ on the vector $({b}-{d}) \times {c}$ is

If $\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c}$ are the pth,qth,nth terms of an HP and $\mathrm{\vec u} = \left( {q - r} \right)\mathbf{\hat i} + \left( {r - p} \right)\mathbf{\hat j} + \left( {p - q} \right)\mathbf{\hat k}$ and $\mathrm{\vec v} = \frac{{\mathbf{\hat i}}}{a} + \frac{{\mathbf{\hat j}}}{b} + \frac{{\mathbf{\hat k}}}{c}$, then

A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$ is normal to $\vec{v}$, then

If $|\mathbf{a}|=5,|\mathbf{a}-\mathbf{b}|=8$ and $|\mathbf{a}+\mathbf{b}|=10$, then $|\mathbf{b}|$ is equal to

The angle between $3 \mathbf{i}+4 \mathbf{j}$ and $2 \mathbf{j}-5 \mathbf{k}$ is

If the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ be $\theta$ and $\mathbf{a} \cdot \mathbf{b}=\cos \theta$, then the true statement is

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

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

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

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?

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