Dot product of vectors practice · Mixed difficulty

$\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c}$ are mutually perpendicular unit vectors, then $\left| {\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}} \right|$ is equal to

Given that $\left| {\,\mathbf{\vec a}\,} \right| = 3,\,\,\left| {\,\mathbf{\vec b}\,\left| { = 4,\,} \right|\,\mathbf{\vec a} \times \mathbf{\vec b}\,} \right| = 10$, then $|\mathbf{\vec a} \cdot \mathbf{\vec b}{|^2}$ equals

If $\vec a = 4\hat i + 6\hat j$ and $\vec b = 3\hat j + 4\hat k$, then the vector form of component of ${\vec a}$ along ${\vec b}$ is

The position vector of a point $P$ is ${r}=x {i}+y {j}+z {k}$, where $x, y, z \in {N}$ and ${u}={i}+{j}+{k}$. If ${r} \cdot {u}=10$, then the number of possible positions of $P$ is

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and let $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is:

If \vec a = \widehat i + 2\widehat j - 3\hat k$\mathbf{\vec a} = \widehat \mathbf{i} + 2\widehat \mathbf{j} - 3\mathbf{\hat k}$ and \vec b = 3\widehat i - \widehat j + 2\hat k,$\mathbf{\vec b} = 3\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k},$ then the angle between the vectors $\mathbf{\vec a} + \mathbf{\vec b}$ and $\mathbf{\vec a} - \mathbf{\vec b}$ is

If $\bar{a}_1, \bar{a}_2, \bar{a}_3, \ldots . \bar{a}_{20}$ are 20 unit vectors such that $\sum\limits_{{i=1}}^{20} \bar{a}_i=0$ then the value of $\sum\limits_{{0 \leq I < j \leq 20}} \overline{a_i} \cdot \overline{a_j}$ is

If the angle between $\mathbf{\hat i} + \mathbf{\hat k}$ and $\mathbf{\hat i} + \mathbf{\hat j} + a\mathbf{\hat k}$ is $\frac{\pi }{3}$, then the value of $a$ is

The value of $x$ for which the angle between the vectors $\mathbf{a}=-3 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}=x \hat{\mathbf{i}}+2 x \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is acute and the angle between $\mathbf{b}$ and $X$-axis lies between $\pi / 2$ and $\pi$ satisfy

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

$\left( {\mathbf{\vec a} \cdot \widehat {\mathbf{\dot i}}} \right)\,\,\widehat {\mathbf{\dot i}} + \left( {\mathbf{\vec a} \cdot \widehat {\mathbf{\dot j}}} \right)\,\,\widehat {\mathbf{\dot j}} + \left( {\mathbf{\vec a} \cdot \mathbf{\hat k}} \right)\,\,\mathbf{\hat k}$̂ is equal to

If ${{\mathbf{\vec u}}_1}\;\mathrm{and\;}{{\mathbf{\vec u}}_2}$ be vectors of unit length and $\mathrm{\theta }$ be the angle between them, then

$\frac{1}{2}\left| {{{\mathbf{\vec u}}_2} - {{\mathbf{\vec u}}_1}} \right|\mathrm{is}$

The projection vector of $\vec{a}$ on $\vec{b}$ is

If $\hat{\mathbf{a}}$ is a unit vector and projection of $\mathbf{x}$ along $\hat{\mathbf{a}}$ is 2 units and $(\hat{\mathbf{a}} \times \mathbf{x})+\mathbf{b}=\mathbf{x}$, then $\mathbf{x}$ 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

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2 \vec{b}}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

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

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

Let $\mathbf{u}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{v}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\mathbf{w}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. If $\mathbf{n}$ is a unit vector such that $\mathbf{u} \cdot \mathbf{n}=0$ and $\mathbf{v} \cdot \mathbf{n}=0$, then $|\mathbf{w} \cdot \mathbf{n}|$ is equal to

$L_1$ 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

A particle moves along a curve so that its coordinates at time $t$ are $x=t, y=\frac{1}{2} t^2, z=\frac{1}{3} t^3$. The acceleration at $t=1$ 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

Let \vec a = \widehat i - \widehat j$\mathbf{\vec a} = \widehat \mathbf{i} - \widehat \mathbf{j}$, \vec b = \widehat j - \hat k, \vec c = \hat k - \widehat i$\mathbf{\vec b} = \widehat \mathbf{j} - \mathbf{\hat k},\;\mathbf{\vec c} = \mathbf{\hat k} - \widehat \mathbf{i}$. If ${\mathbf{\vec d}}$ is a unit vectors such that $\mathbf{\vec a}\cdot\mathrm{\;}\mathbf{\vec d} = 0 = \left[ {\mathrm{\;}\mathbf{\vec b}\;\mathrm{\;}\mathbf{\vec c}\;\mathrm{\;}\mathbf{\vec d}} \right],$ then ${\mathbf{\vec d}}$ is (are)

If vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $|\vec{a}-\vec{c}|=|\vec{b}-\vec{c}|$, then $(\vec{b}-\vec{a}) \cdot\left(\vec{c}-\frac{\vec{a}+\vec{b}}{2}\right)$ 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 $\vec c = x\vec a + y\vec b + \vec c\left( {\vec a \times \vec b} \right)$, then

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}+4 \hat{k}$ be two vectors, such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{i}+12 \hat{k}$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to

Let $\mathbf{\vec a},\mathbf{\vec b}$ and $\mathbf{\vec c}\;$ be vectors with magnitude 3,4 and 5 respectively and $\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c} = \vec 0$ , then the value of $\mathbf{\vec a}\cdot\mathbf{\vec b} + \mathbf{\vec b}\cdot\mathbf{\vec c} + \mathbf{\vec c}\cdot\mathbf{\vec a}$ 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

Two forces $\overrightarrow{{P}}=3 \hat{{i}}-2 \hat{{j}}+\hat{{k}}$ and $\overrightarrow{{Q}}=\hat{{i}}+3 \hat{{j}}-5 \hat{{k}}$ acting on a particle $A$ move it to $B$. If the position vectors of $A$ and $B$ are respectively $-2 \hat{{i}}+5 \hat{{k}}$ and $3 \hat{{i}}-7 \hat{{j}}+2 \hat{{k}}$, then the total work done is

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors, such that $\mathbf{a}+\mathbf{b}+\mathbf{c}=0,|\mathbf{a}|=1$, $|\mathbf{b}|=2,|\mathbf{c}|=3$, then $\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}$ is equal to