Addition of vectors practice · Mixed difficulty

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be any three non-coplanar vectors, then $[a+b ~~~b+c~~~~ c+a]=$

Which of the following is/are true?

I. In a zero vector, initial and terminal points coincide.

II. Zero vector is denoted as $\mathrm{O}$.

III. Zero vector has zero magnitude.

The value of $\left[ {\vec a - \vec b,\vec b - \vec c,\vec c - \vec a} \right],$ where $\left| {\vec a} \right| = 1,\left| {\vec b} \right| = 5,\left| {\vec c} \right| = 3,$ is

The vector \vec a = α \widehat i + 2\widehat j + β \hat k$\mathbf{\vec a} = \mathrm{\alpha }\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathrm{\beta }\mathbf{\hat k}$ lies in the plane of the vectors \vec b = \widehat i + \widehat j$\mathbf{\vec b} = \widehat \mathbf{i} + \widehat \mathbf{j}$ and \vec c = \widehat j + \hat k$\vec c = \widehat \mathbf{j} + \mathbf{\hat k}$

and bisects the angle between ${\mathbf{\vec b}}$ and ${\mathbf{\vec c}}$ Then, which one of the following gives possible value of $\mathrm{\alpha }$ and $\mathrm{\beta }$?

If vectors $3\hat i + 2\hat j - \hat k\,\,\,\,{\text{and}}\,\,\,6\hat i - 4x\hat j + y\hat k$ are parallel, then the value of x and y will be

In a trapezium, the vector $BC = \lambda AD.$ We will then find that p = AC + BD is collinear with AD. If $p = \mu AD,$ then

The position vectors of the points $A,B,C$ are $\left( {2\mathbf{\hat i} + \mathbf{\hat j} - \mathbf{\hat k}} \right),\mathrm{\;}\left( {3\mathbf{\hat i} - 2\mathbf{\hat j} + \mathbf{\hat k}} \right)$ and $\left( {\mathbf{\hat i} + 4\mathbf{\hat j} - 3\mathbf{\hat k}} \right)$ respectively. These points

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\vec{a}+2 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3 \vec{c}$ is collinear with $\vec{a}$ ( $\lambda$ being some non-zero scalar), then $\vec{a}+2 \vec{b}+6 \vec{c}$ equals

If three points $A, B$ and $C$ have position vectors $(1, x, 3),(3,4,7)$ and $(y,-2,-5)$ respectively and if they are collinear, then $(x, y)=$

The vectors a, b and a + b are

If $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}, \vec{c}=3 \hat{i}+\hat{j}+2 \hat{k}$ and $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=-3(\hat{i}-\hat{k})$, then the number triple $(\alpha, \beta, \gamma)$ is

Five points given by $A,B,C,D,E$ are in plane. Three forces $\overrightarrow {\mathbf{AC}} ,\;\overrightarrow {\;\mathbf{AD}}$ and $\overrightarrow {\mathbf{AE}}$ act a $A$ and three forces $\overrightarrow {\mathbf{CB}} ,\overrightarrow {\mathbf{DB}} ,\overrightarrow {\mathbf{EB}}$ act at $B$. Then, their resultant is

$A$ vector a has the components $2 p$ and 1 w.r.t. a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If with respect to a new system, a has components $(p+1)$ and 1 , then

A unit vector a makes an angle $\frac{\pi }{4}$ with Z – axis. If $a = \hat i + \hat j$ is a unit vector, then a is equal to

The position vectors of A and B are $2\hat i - 9\hat j - 4\hat k\,\,\,{\text{and}}\,\,6\hat i - 3\hat j + 8\hat k$ respectively, then the magnitude of AB is

If $a = \hat i + \hat j + \hat k,b = 4\hat i + 3\hat j + 4\hat k\,\,\,\,{\text{and}}\,\,\,\,c = \hat i + \alpha \hat j + \beta \hat k$ are linearly dependent vectors and $\left| c \right| = \sqrt 3$ then

If $O$ is origin of $C$ is the mid point of $A\left( {2, - 1} \right)$ and $B\left( { - 4,\;3} \right)$. Then, the value of $\overrightarrow {\mathbf{OC}}$ is

In a quadrilateral $P Q R S, \mathbf{P Q}=a, \mathbf{Q R}=b$, $\mathbf{S P}=a-b$. IfM is the mid-point of $Q R$ and $X$ is a point of $S M$ such that, $S X=\frac{4}{5} S M$, then

Vectors $\vec{a}$ and $\vec{b}$ are non-zero and non-collinear. The value of $x$ for which $\vec{c}=(x-2) \vec{a}+\vec{b}$ and $\bar{d}=(2 x+1) \vec{a}-\vec{b}$ are collinear, is

If $\bar{a}, \bar{b}, \bar{c}$ are unit vectors such that $\bar{a}$ is perpendicular to plane of $\bar{b}$ and $\bar{c}$ and the angle between $\bar{b} \& \bar{c}$ is $\frac{\pi}{3}$ the $|\bar{a}+\bar{b}+\bar{c}|$ is

Let $P(3,2,6)$ be a point in space and $Q$ be a point on the line $\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})$. Then the value of $\mu$ for which the vector $P Q$ is parallel to the plane $x-4 y+3 z=1$ is

If $(1,\,5,\,35),\,(7,\,5,\,5),\,\left( {1,\,\lambda ,\,7} \right)$ and $(2\lambda ,\,1,\,2)$ are coplanar, then the sum of all possible values of $\lambda$ is

If the points with position vectors 60\widehat i + 3\widehat j,40\widehat i - 8\widehat j$60\widehat \mathbf{i} + 3\widehat \mathbf{j},40\widehat \mathbf{i} - 8\widehat \mathbf{j}$ and a\widehat i - 52\widehat j$a\widehat \mathbf{i} - 52\widehat \mathbf{j}$ are collinear, then $a$ is equal to

Find the value of $\lambda$ so that the points P, Q, R and S on the sides OA, OB, OC and AB, respectively, of a regular tetrahedron OABC are coplanar. It is given that $\frac{{OP}}{{OA}} = \frac{1}{3},\,\,\,\,\frac{{OQ}}{{OB}} = \frac{1}{2},\,\,\,\,\,\frac{{OR}}{{OC}} = \frac{1}{3}\,\,\,{\text{ and }}\,\,\frac{{OS}}{{AB}} = \lambda .$

Let a, b and c be three unit vectors such that 3a + 4b + 5c = 0. Then which of the following statements is true?

The vector \vec a = \widehat i + \widehat j + m\hat k$\mathbf{\vec a} = \widehat \mathbf{i} + \widehat \mathbf{j} + m\mathbf{\hat k}$, \vec b = \widehat i + \widehat j + ( m + 1 )\hat k$\mathbf{\vec b} = \widehat \mathbf{i} + \widehat \mathbf{j} + \left( {m + 1} \right)\mathbf{\hat k}$ and \vec c = \widehat i - \widehat j + m\hat k$\mathbf{\vec c} = \widehat \mathbf{i} - \widehat \mathbf{j} + m\mathbf{\hat k}$ are coplanar, if $m$ is equal to

The position vector of a point C with respect to B is $\hat i + \hat j$ and that of B with respect to A is $\hat i - \hat j.$ The position vector of C with respect to A is

If a and b are two non – collinear vectors xa + yb = 0

If $|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5$ and $\vec{a}+\vec{b}+\vec{c}=0$, then the angle between $\vec{a}$ and $\vec{b}$ is:

$P$ is the point of intersection of the diagonals of the parallelogram $A B C D$. If $O$ is any point, then $\mathbf{O A}+\mathbf{O B}+\mathbf{O C}+\mathbf{O D}$ is equal to