Cross product practice · Easy

Let $|\mathbf{a}|=|\mathbf{b}|=2$ and $\mathbf{p}=\mathbf{a}+\mathbf{b}, \mathbf{q}=\mathbf{a}-\mathbf{b}$. If $|\mathbf{p} \times \mathbf{q}|=2\left(k-(\mathbf{a} \cdot \mathbf{b})^2\right)^{1 / 2}$ then

If $\vec u = \vec a - \vec b,\;\vec v = \vec a + \vec b$ and $\left| {\vec a} \right| = \left| {\vec b} \right| = 2$, then $\left| {\vec u \times \vec v} \right|$ is

The area of a triangle whose vertices are $A(1,-1,2)$, $B(2,1,-1)$ and $C(3,-1,2)$ is

Consider a tetrahedron with faces ${F}_1, {~F}_2, {~F}_3, {~F}_4$. Let ${V}_1, {V}_2, {V}_3, {V}_4$ be the vectors whose magnitudes are respectively equal to areas of ${F}_1, {~F}_2, {~F}_3, {~F}_4$ and whose directions are perpendicular to their faces in outward direction. Then $\left|{V}_1+{V}_2+{V}_3+{V}_4\right|$ equals

A vector of magnitude 12 units perpendicular to the plane containing the vectors 4\widehat i + 6\widehat j - \hat k$4\widehat \mathbf{i} + 6\widehat \mathbf{j} - \mathbf{\hat k}$ and 3\widehat i + 8\widehat j + \hat k$3\widehat \mathbf{i} + 8\widehat \mathbf{j} + \mathbf{\hat k}$ is

If $\vec a,\;\vec b,\;\vec c$ are non-collinear vectors such that $\vec a + \vec b$ is parallel to ${\vec c}$ and $\vec c + \vec a$ is parallel to ${\vec b}$, then

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} . \vec{b}=1$ and $\vec{a} \times \vec{b}=\vec{j}-\vec{k}$, then $\vec{b}$ is

If $\vec{a}, \vec{b}$, and $\overrightarrow{\mathrm{c}}$ are unit vectors such that $\vec{a}+2 \vec{b}+2 \overrightarrow{\mathrm{c}}=\overrightarrow{0}$, then $|\vec{a} \times \overrightarrow{\mathrm{c}}|$ is equal to

Let $a = 3\hat i + 2\hat j + 2\hat k$ and $b = \hat i + 2\hat j + 2\hat k$ be two vectors. If a vector perpendicular to both the vectors a+b and a-b has the magnitude 12, then one such vector is

Let $A_0 A_1 A_2 A_3 A_4 A_5$ be a regular hexagon with vertex $A_0$ and $A_3$ at $(1,0)$ and $(-1,0)$ respectively. The equations representing $A_1 A_4$ and $A_2 A_5$ are

Given $|\vec{b}|=2,|\vec{b} \times \vec{a}|=2$

Then $|\vec{b} \times \vec{a}-\vec{b}|^2$ is

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three coplanar vectors, then $[a+b b+c c+a]=$

If $\overrightarrow{F_1}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \quad \overrightarrow{F_2}=-\mathbf{i}+2 \mathbf{j}-\mathbf{k}, \quad \overrightarrow{F_3}=\mathbf{j}-\mathbf{k}$, $\vec{A}=4 \mathbf{i}-3 \mathbf{j}-2 \mathbf{k}$ and $\vec{B}=6 \mathbf{i}+\mathbf{j}-3 \mathbf{k}$, then the scalar product of $\vec{F}_1+\vec{F}_2+\vec{F}_3$ and $\overrightarrow{A B}$ will be

$|\vec{a} \times \hat{i}|^2+|\vec{a} \times \hat{j}|^2+|\vec{a} \times \hat{k}|^2=$

The sine of the angle between the two vectors $3 \mathbf{i}+2 \mathbf{j}-\mathbf{k}$ and $12 \mathbf{i}+5 \mathbf{j}-5 \mathbf{k}$ will be

If $|\vec{a}|=4,|\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi / 6$ then $(\vec{a} \times \vec{b})^2$ is equal to

A unit vector perpendicular to the plane of $\mathrm{\vec a} = 2\mathrm{\hat i} - 6\mathrm{\hat j} - 3\mathrm{\hat k},\mathrm{\;\vec b} = 4\mathrm{\hat i} + 3\mathrm{\hat j} - \mathrm{\hat k}$ is

The area of the triangle formed by $A(1,0,0)$, $B(0,1,0), C(1,1,1)$ is

The moment about the point $M\left( { - 2,\;4, - 6} \right)$ of the force represented in magnitude and position $\overrightarrow {\mathbf{AB}}$ where the points $A$ and $B$ have the coordinates (1, 2,$- 3)$ and $\left( {3, - 4,2} \right)$ respectively, is

If $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$, then $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ is equal to

If $\mathbf{\vec a} = \mathbf{\hat i} + 2\mathbf{\hat j} + 3\mathbf{\hat k}\mathrm{\;and\;}\mathbf{\vec b} = \mathbf{\hat i} \times \left( {\mathbf{\vec a} \times \mathbf{\hat i}} \right) + \mathbf{\hat j} \times \left( {\mathbf{\vec a} \times \mathbf{\hat j}} \right) + \mathbf{\hat k} \times \left( {\mathbf{\vec a} \times \mathbf{\hat k}} \right)$, then length of ${\mathbf{\vec b}}$ is equal to

Consider the parallelopipped with sides a $=3 {i}+2 {j}+{k}, {b}={i}+{j}+2 {k}$ and ${c}={i}+3 {j}+3 {k}$ then the angle between $a$ and the plane containing the face determined by ${b}$ and ${c}$ is

$|\mathbf{a} \times \mathbf{i}|^2+|\mathbf{a} \times \mathbf{j}|^2+|\mathbf{a} \times \mathbf{k}|^2=$

If $3 \mathbf{i}+4 \mathbf{j}$ and $-5 \mathbf{i}+7 \mathbf{j}$ are the vector sides of any triangle, then its area is given by

Area of rhombus is ……., where diagonals are \vec a = 2\widehat i - 3\widehat j + 5\hat k$\mathbf{\vec a} = 2\widehat \mathbf{i} - 3\widehat \mathbf{j} + 5\mathbf{\hat k}$ and \vec b = - \widehat i + \widehat j + \hat k$\mathbf{\vec b} = - \widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}$

For any vector ${\mathbf{\vec a}}$, the value of (\vec a × \widehat i)² + (\vec a × \widehat j)² + (\vec a × \hat k)²${(\mathbf{\vec a} \times \widehat \mathbf{i})^2} + {(\mathbf{\vec a} \times \widehat \mathbf{j})^2} + {(\mathbf{\vec a} \times \mathbf{\hat k})^2}$ is equal to

If $|\vec{a}|=2$ and $|\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=0$, then $((\vec{a} \times(\vec{a} \times(\vec{a} \times(\vec{a} \times \vec{b})))$ is equal to

If $\overrightarrow{O A}=3 \mathbf{i}+2 \mathbf{j}-\mathbf{k}$ and $\overrightarrow{O B}=i+3 \mathbf{j}+\mathbf{k}$, then the area of the triangle $O A B$ is

Let $a = 2\hat i + \hat j - 2\hat k$ and $b = \hat i + \hat j$ , if c is a vector such that $a.\,c = \left| c \right|,\left| {c - a} \right| = 2\sqrt 2$ and the angle between $\left( {a \times b} \right)$ and c is $\frac{\pi }{6},$ then the value of $\left| {\left( {a \times b} \right) \times c} \right|$ is

If $\mathbf{a}=2 \mathbf{i}-3 \mathbf{j}-\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+4 \mathbf{j}-2 \mathbf{k}$, then $\mathbf{a} \times \mathbf{b}$ is