Dot product of vectors practice · Medium

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$. If a vector $\vec{d}$ satisfies $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{d} \cdot \vec{a}=24$, then $|\vec{d}|^{2}$ is equal to

Let $\overrightarrow a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\,\,\,{a_i} > 0,\,\,i = 1,2,3\,$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\overrightarrow a$ on the vector $3\hat i + 4\hat j$ be 7. Let $\overrightarrow b$ be a vector obtained by rotating $\overrightarrow a$ with $90^\circ .$ If $\overrightarrow a .\,\overrightarrow b$ and x – axis are coplanar, then the projection of a vector $\overrightarrow b$ on $3\hat i + 4\hat j$ is equal to

If the projection of the vector $\mathbf{\vec a}\;\mathrm{on\;}\mathbf{\vec b}$ is $\left| {\mathbf{\vec a} \times \mathbf{\vec b}} \right|$ and if 3\vec b = \widehat i + \widehat j + \hat k$3\mathbf{\vec b} = \widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}$, then the angle between $\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b}$ is

$L_1$ and $L_2$ are two lines whose vector equations are
$L_1: \mathbf{r}=\lambda[(\cos \theta+\sqrt{3}) \mathbf{i}+(\sqrt{2} \sin \theta) \mathbf{j}+(\cos \theta-\sqrt{3}) \mathbf{k}]$
and $L_2: \mathbf{r}=\mu(a \mathbf{i}+b \mathbf{j}+c \mathbf{k})$
where, $\lambda$ and $\mu$ are scalars and $\alpha$ is the acute angle between $L_1$ and $L_2$. If the angle $\alpha$ is independent of $\theta$, then the value of $\alpha$ is

Forces of magnitudes 3 and 4 units acting along 6\widehat i + 2\widehat j + 3\hat k$6\widehat \mathbf{i} + 2\widehat \mathbf{j} + 3\mathbf{\hat k}$ and 3\widehat i - 2\widehat j + 6\hat k$3\widehat \mathbf{i} - 2\widehat \mathbf{j} + 6\mathbf{\hat k}$ respectively act on a particle and displace it from $\left( {2,2 - 1} \right)$ to $\left( {4,3,1} \right)$. The work done is

If $a, b, c$ be non-zero vectors such that a is perpendicular to $\mathbf{b}$ and $\mathbf{c}$ and $|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{c}|=1, \mathbf{b} \cdot \mathbf{c}=1$ and there is a non-zero vector $\mathbf{d}$ coplanar with $\mathbf{a}+\mathbf{b}$ and $2 \mathbf{b}-\mathbf{c}$ and $\mathbf{d} \cdot \mathbf{a}=1$, then minimum value of $|\mathbf{d}|$ is

The values of $c$ so that for all real $x$, the vectors $c x {i}-6 {j}+3 {k}$ and $x {i}+2 {j}+2 c x {k}$ make an obtuse angle with each other, are

If the equation of the plane passing through the point $(-1,2,0)$ and parallel to the lines $\frac{x}{3}=\frac{y+1}{0}=\frac{z-2}{-1}$ and $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-1}$ is $a x+b y+c z=1$, then the value of $(a+b+c)$, is

The unit vector in $ZOX$ plane and making angle $45^\circ$ and $60^\circ$ respectively with\vec a = 2\widehat i + 2\widehat j - \hat k and \vec b = 0\widehat i + \widehat j - \hat k,$\mathbf{\vec a} = 2\widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}\;\mathrm{and}\;\mathbf{\vec b} = 0\widehat \mathbf{i} + \widehat \mathbf{j} - \mathbf{\hat k},$ is

The equation of the plane passing through the line of intersection of the planes
$r.(\hat i + \hat j + \hat k) = 1$ and $r.(2\hat i + 3\hat j - \hat k) + 4 = 0$ and parallel to the $X$-axis is

The distance of a point $(7,-3,-4)$ from the plane passing through the points $(2,-3,1),(-1,1,-2)$ and $(3,-4,2)$ is

The projection of $2 \hat{i}-3 \hat{j}+4 \hat{k}$ on the line whose equation is $\vec{r}=(3+\lambda) \hat{i}+(3-2 \lambda) \hat{j}+(5+6 \lambda) \hat{k}$ where $\lambda$ is a scalar parameter, is

If two adjacent sides of two rectangles are represented by the vectors $\mathbf{p}=5 \mathbf{a}-3 \mathbf{b}, \mathbf{q}=-\mathbf{a}-2 \mathbf{b}$ and $r=-4 a-b ; s=-a+b$ respectively, then the angle between the vectors $\mathbf{x}=\frac{1}{3}(\mathrm{p}+\mathbf{r}+s)$ and $\mathrm{y}=\frac{1}{5}(\mathbf{r}+s)$

Two forces whose magnitudes are $2 \mathrm{gm}$ wt, and $3 \mathrm{gm}$ wt act on a particle in the directions of the vectors $2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $4 \hat{i}+4 \hat{j}+2 \hat{k}$ resepectively. If the particle is displaced from the origin to the point $(1,2,2)$, the work done is (the unit of length being $1 \mathrm{~cm}$ ) :

If $l \mathbf{a}+m \mathbf{b}+n \mathbf{c}=\mathbf{0}$, where $l, m, n$ are scalars and $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are mutually perpendicular vectors, then

Let $L$ be the line of intersection of planes $r.(\hat i - \hat j + 2\hat k) = 2$ and $r.(2\hat i + \hat j - \hat k) = 2$. If $P(\alpha ,\,\beta ,\,\gamma )$ is the foot of perpendicular on $L$ from the point (1, 2, 0), then the value of $35(\alpha + \,\beta + \,\gamma )$ is equal to

Equation of the straight line in the plane $\vec{r}. \vec{n}=d$ which is parallel to $\vec{r}=\vec{a}+\lambda \vec{b}$ and passes through the foot of perpendicular drawn from the point $P(\vec{a})$ to the plane $\vec{r} .\vec{n}=d$ is (where $\vec{n} \cdot \vec{b}=0)$

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{{C}}$. Then $\overrightarrow{{C}}+\left(-\frac{1}{2} \hat{{i}}+\frac{1}{3 \sqrt{2}} \hat{{j}}-\frac{\sqrt{2}}{3} \hat{{k}}\right)$ is:

Let $\vec{a}=\hat{{i}}+2 \hat{{j}}-3 \widehat{{k}}$ and $\vec{b}=2 \hat{{i}}-3 \hat{{j}}+5 \widehat{{k}}$. If $\vec{r} \times \vec{a}=\vec{b} \times \vec{r}, \, \vec{r} \cdot(\alpha \hat{{i}}+2 \hat{{j}}+\hat{{k}})=3$ and $\vec{r} \cdot(2 \hat{{i}}+5 \hat{{j}}-\alpha \hat{{k}})=-1, \alpha \in R$, then the value of $\alpha+|\vec{r}|^2$ is equal to:

Let a,b and c be three vectors such that $\mathrm{a}\,\mathrm{ = }\,\mathrm{b \times }\,\mathrm{(b \times c)}\mathrm{.}$ . If magnitude of the vectors a, b and c are $\sqrt 2 ,\,\,1\,\,\mathrm{and}\,\,2$, respectively and the angle between b and c is $\theta \,\,\left( {0 < \theta < \frac{\pi }{2}} \right)\,$, then the Value of $(1 + \tan \theta )$ 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

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,

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

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