Dot product of vectors practice · Mixed difficulty

If the position vectors of the points $A$ and $B$ are $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ respectively, then the equation of the plane through $B$ and perpendicular to $A B$ is

Vectors $\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b}$ are inclined at an angle $\mathrm{\theta } = 120^\circ$. If $\left| {\mathbf{\vec a}} \right| = 1,\;\left| {\mathbf{\vec b}} \right| = 2,$ then $[\left( {\mathbf{\vec a} + 3\mathbf{\vec b}} \right) \times {(3\mathbf{\vec a} + \mathbf{\vec b}]^2}$ is equal to

Let a vector $\vec{a}$ has magnitude 9 . Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$, the vector $(x \vec{a}+y \vec{b})$ is perpendicular to the vector $(6 y \vec{a}-18 \times \vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to

If the position vectors of the points $A, B, C, D$ be $\mathbf{i}+\mathbf{j}+\mathbf{k}, 2 \mathbf{i}+5 \mathbf{j}, 3 \mathbf{i}+2 \mathbf{j}-3 \mathbf{k}$ and $\mathbf{i}-6 \mathbf{j}-\mathbf{k}$, then the angle between the vectors $\overrightarrow{A B}$ and $\overrightarrow{C D}$ is

If the vector 3\widehat i - 2\widehat j - 5\hat k$3\widehat \mathbf{i} - 2\widehat \mathbf{j} - 5\mathbf{\hat k}$ is perpendicular to c\hat k - \widehat j + 6\widehat i$c\mathbf{\hat k} - \widehat \mathbf{j} + 6\widehat \mathbf{i}$ then $c$ is equal to

The cartesian from of the plane $\mathbf{\vec r} = \left( {s - 2t} \right)\widehat {\mathbf{\dot i}} + \left( {3 - t} \right)\widehat {\mathbf{\dot j}} + \left( {2s + t} \right)\mathbf{\hat k}$ is

Let $|\mathbf{a}|=3$ and $|\mathbf{b}|=4$. The value of $\mu$ for which the vectors $\mathbf{a}+\mu \mathbf{b}$ and $\mathbf{a}-\mu \mathbf{b}$ will be perpendicular is

If $\mathbf{\hat u}\;\mathrm{and\;}\mathbf{\hat v}$ unit vectors and $\mathrm{\theta }$ is the acute angle between them, then $2\mathbf{\hat u} \times 3\mathbf{\hat v}$ is a unit vector for

Let $a = \hat i - 2\hat j + \hat k\,\,\mathrm{and}\,\,b = \hat i - \hat j + \hat k$ be two vectors. If $c$ is a vector such that $b \times c = \,b \times a\,\,\mathrm{and}\,\,c.\,a = 0$ then $c \cdot b$ is equal to

The vector equation of the plane through the point $(2,1,-1)$ and passing through the line of intersection of the plane $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0$ and $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=0$ is -

If ${a}, {b}, {c}$ are unit vectors such that ${a}-{b}+{c}={0}$ then ${c} \cdot {a}$ is equal to

The distance between the line \vec r = 2\widehat i - 2\widehat j + 3\hat k + λ ( \widehat i - \widehat j + \hat k )$\mathbf{\vec r} = 2\widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\mathbf{\hat k} + \lambda \;\left( {\widehat \mathbf{i} - \widehat \mathbf{j} + \mathbf{\hat k}} \right)$ and the plane \vec r·( \widehat i + 5\widehat j + \hat k ) = 5$\mathbf{\vec r}\cdot\left( {\widehat \mathbf{i} + 5\widehat \mathbf{j} + \mathbf{\hat k}} \right) = 5$ is

Let two non-collinear vectors $\mathbf{a}$ and $\mathbf{b}$ inclined at an angle $\frac{2 \pi}{3}$ be such that $|\mathbf{a}|=3$ and $|\mathbf{b}|=2$. If a point $P$ moves so that at any time $t$ its position vector $\mathrm{OP}$ (where $O$ is the origin) is given as $\mathbf{O P}=\left(t+\frac{1}{t}\right) \mathbf{a}+\left(t-\frac{1}{t}\right) \mathbf{b}$, then least distance of $P$ from the origin is

The vector equation of the line passing through the points $\left( {3,2,1} \right)$ and $\left( { - 2,1,3} \right)\;$ is

The point of intersection of the plane $\vec{r} \cdot(3 \hat{i}-5 \hat{j}+2 \hat{k})=6$ with the straight line passing through the origin and perpendicular to the plane $2 x-y-z=4$, is $\left(x_3, y_0, z_0\right)$. The value of $\left(2 x_0-3 y_0+z_0\right)$, is:

If $\vec{x}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{y}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\vec{z}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}$,
then the magnitude of the projection of $\vec{x} \times \vec{y}$ on $\vec{z}$ is:

The angle between a diagonal of a cube and one of its edges is

Let $\mathbf{a}$ and $\mathbf{b}$ be two unit vectors. If the vectors $\mathbf{c}=\mathbf{a}+2 \mathbf{b}$ and $\mathbf{d}=5 \mathrm{a}-4 \mathrm{a}$ are perpendicular to each other, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is

If the position vectors of $A,B\;\mathrm{and\;}C$ are respectively 2\widehat i - \widehat j + \hat k, \widehat i - 3\widehat j - 5\hat k and 3\widehat i - 4\dot Ĵ - 4\hat k $2\widehat \mathbf{i} - \widehat \mathbf{j} + \mathbf{\hat k},\mathrm{\;}\widehat \mathbf{i} - 3\widehat \mathbf{j} - 5\mathbf{\hat k}\mathrm{\;and\;}3\widehat \mathbf{i} - 4\widehat {\mathbf{\dot J}} - 4\mathbf{\hat k}\mathrm{\;}$ then $\mathrm{co}{\mathrm{s}^2}\;A$ is equal to

If $\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec a + \vec b} \right| = 1$, then $\left| {\vec a - \vec b} \right|$ is equal to

Let $\overrightarrow a$ and $\overrightarrow b$ be the vectors along the diagonals of a parallelogram having area $2\sqrt 2 .$ Let the angle between $\overrightarrow a$ and $\overrightarrow b$ be acute, $\left| {\overrightarrow a } \right| = 1,$ and $\left| {\overrightarrow a \cdot \overrightarrow b } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|.$ If $\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b ,$ then an angle between $\overrightarrow b$ and $\overrightarrow c$ is

Let $a$ and $b$ be two vectors such that $\left| {\mathrm{2a + 3b}} \right|\mathrm{ = }\,\left| {\mathrm{3a + b}} \right|$ and the angle between a and b is $60^\circ .$ If $\frac{1}{8}a$ is a unit vector, then $\left| \mathrm{b} \right|$ is equal to

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{4}$. If $\theta$ is the angle between the vectors $(\hat{a}+\hat{b})$ and $(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$, then the value of $164 \cos ^2 \theta$ is equal to :

The total work done by two forces \vec F₁ = 2\widehat i - \widehat j${{\mathbf{\vec F}}_1} = 2\widehat \mathbf{i} - \widehat \mathbf{j}$ and \vec F₂ = 3\widehat i + 2\widehat j - \hat k${{\mathbf{\vec F}}_2} = 3\widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}$ acting on a particle when it is displaced from the point 3\widehat i + 2\widehat j + \hat k$3\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k}$ to 5\widehat i + 5\widehat j + 3\hat k$5\widehat \mathbf{i} + 5\widehat \mathbf{j} + 3\mathbf{\hat k}$ is

If the vectors $3 \mathbf{i}+\lambda \mathbf{j}+\mathbf{k}$ and $2 \mathbf{i}-\mathbf{j}+8 \mathbf{k}$ are perpendicular, then $\lambda$ is

If $\mathbf{a} \cdot \mathbf{b}$ and $\mathbf{c}$ are any three non-zero vectors, then the component of $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ perpendicular to $\mathrm{b}$ is

Let $\vec{a}=\alpha \hat{i}+\hat{j}-k$ and $\vec{b}=2 \hat{i}+\hat{j}-\alpha k, \alpha>0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $-\hat{i}+2 \hat{j}-2 k$ is 30 , then $\alpha$ is equal to

A parallelogram is constructed on $3\mathbf{\vec a} + \mathbf{\vec b}\mathrm{\;and\;}\mathbf{\vec a} - 4\mathbf{\vec b}$, where $\left| {\mathbf{\vec a}\left| { = 6\mathrm{\;and\;}} \right|\mathbf{\vec b}} \right| = 8\mathrm{\;and\;}\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b}$ are anti-parallel, then the length of the longer diagonal is

The vector equation of the plane containing the lines $\mathbf{r}=(\mathbf{i}+\mathbf{j})+\lambda(\mathbf{i}+2 \mathbf{j}-\mathbf{k})$ and $\mathbf{r}=(\mathbf{i}+\mathbf{j})+\mu(-\mathbf{i}+\mathbf{j}-2 \mathbf{k})$ is

The direction cosines of the resultant of the vectors $(\mathbf{i}+\mathbf{j}+\mathbf{k}),(-\mathbf{i}+\mathbf{j}+\mathbf{k}),(\mathbf{i}-\mathbf{j}+\mathbf{k})$ and $(\mathbf{i}+\mathbf{j}-\mathbf{k})$, are