Addition of vectors practice · Mixed difficulty

If $A,B,C$ are vertices of a triangle whose position vectors are $\vec a,\vec b$ and ${\vec c}$ respectively and $G$ is the centroid of $\mathrm{\Delta }ABC$, then $\vec GA + \vec GB + \vec GC$, is

If ${m_1},\,\,{m_2},\,\,{m_3}\;\mathrm{and\;}{m_4}$ are respectively the magnitudes of the vectors

$\overrightarrow {{\mathbf{a}_1}} = 2\widehat {\,\mathbf{\dot i}} - \widehat {\mathbf{\dot j}} + \mathbf{\hat k},\;\overrightarrow {\,{\mathbf{a}_2}} = 3\widehat {\mathbf{\dot i}} - 4\widehat {\mathbf{\dot j}} + 4\mathbf{\hat k},\;\overrightarrow {\,{\mathbf{a}_3}} = \widehat {\mathbf{\dot i}} + \widehat {\mathbf{\dot j}} - \mathbf{\hat k}\,\;\mathrm{and \;}\overrightarrow {{\mathbf{a}_4}} = \widehat { - \mathbf{\dot i}} + 3\widehat {\,\mathbf{\dot j}} + \mathbf{\hat k}$̂, then the correct order of ${m_1},\,\,{m_2},\,\,{m_3}\,\;\mathrm{and}\,\mathrm{\;}{m_4}$ is

If a and b are two non-zero and non-collinear vectors, then $a + b$ and $a - b$ are

Let two non-collinear unit vectors ${\mathbf{\hat a}}$ and ${\mathbf{\hat b}}$ from and acute angle. A point $P$ moves so that at any time $t$ the position vector $\overrightarrow {\mathbf{OP}}$ (where $O$ is the origin) is given by $\mathbf{\hat a}\cos t + \mathbf{\hat b}\sin t$. When $P$ is farthest form origin $O$, let $M$ be the length of $\overrightarrow {\mathbf{OP}}$ and ${\mathbf{\hat u}}$ be the unit vector along $\overrightarrow {\mathbf{OP}}$ Then,

The direction cosines of the vector $3\hat i - 4\hat j + 5\hat k$ are

If the vectors $\overrightarrow {{r_1}} = a\hat i + \hat j + \hat k,\overrightarrow {{r_2}} = \hat i + b\hat j + \hat k,\overrightarrow {{r_3}} = \hat i + \hat j + c\hat k\left( {a \ne 1,b \ne 1,c \ne 1} \right)$ are coplanar, then the value of $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}},$ is

Let $ABC$ be a triangle the position vectors of whose vertices are respectively $7\hat j + 10\hat k,\; - \hat i + 6\hat j + 6\hat k$ and $- 4\hat i + 9\hat j + 6\hat k$. Then, $\Delta ABC$ is

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

If the position vectors of two point $P$ and $Q$ are respectively $9 \mathbf{i}-\mathbf{j}+5 \mathbf{k}$ and $\mathbf{i}+3 \mathbf{j}+5 \mathbf{k}$, and the line segment $P Q$ intersects the $Y O Z$ plane at a point $R$, the $P R: R Q$ is equal to

If $a = 2\widehat i + \widehat j - 8\widehat k$ and $b = \widehat i + 3\widehat j - 4\widehat k,$ then the magnitude of $a + b$ is equal to

If position vectors of points A, B and C are respectively $\hat i,\,\,\hat j\,\,{\text{and}}\,\,\hat k$ and AB = CX, then position vector of point X is

The number of integral values of p for which $(p + 1)\widehat i - 3\widehat j + p\widehat i,p\widehat i + (p + 1)\widehat j - 3\widehat k$ and $- 3\widehat i + p\widehat j + (p + 1)\widehat k$ are linearly dependent vectors is

A, B, C and D have position vectors a, b, c and d, respectively, such that $a - b = 2(d - c).$ Then,

The direction cosines of the vector $3\widehat i - 4\widehat j + 5\widehat k$ are

Let $a = (1,1 - 1),b = (5, - 3, - 3)$ and $c(3, - 1,2)$ If r is collinear with c and has length $\frac{{\left| {a + b} \right|}}{2}$ then r is equals

Let ${P_1},\,\,{P_2},.....{P_{15}}$ be 15 points on a circle. The number of distinct triangles formed by points ${P_{\hat i}}\,\mathrm{,}{P_{\hat j}}\mathrm{ }\,\,\mathrm{and }\,\,{P_{\hat k}}$ such that $\hat i + \hat j + \hat k \ne 15$ is

If $|\mathbf{a}|=3,|\mathbf{b}|=4$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ be $120^{\circ}$, then $|4 a+3 b|=$

The vector $2 \mathbf{i}+\mathbf{j}-\mathbf{k}$ is perpendicular to $\mathbf{i}-4 \mathbf{j}+\lambda \mathbf{k}$, if $\lambda=$

If $ABCD$ is a quadrilateral, then $\vec BA + \vec BC + \vec CD + \vec DA =$

$A,\;B$ have position vectors $\vec a,\;\vec b$ relative to the origin $O$ and $X,\;Y$ divide $\overrightarrow {AB}$ internally and externally respectively in the ratio 2 : 1. Then, $\overrightarrow {XY} =$

In a $\mathrm{\Delta }ABC$, if $\vec AB = 3\hat i + 4\hat k,\;\vec AC = 5\hat i + 2\hat j + 4\hat k$, then the length of median through $A$, is

The position vector of A and B are $\hat j - \hat j + 2\hat k\,\,\,\,{\text{and}}\,\,\,\,\,\,3\hat i - \hat j + 3\hat k.$ The position vector of the middle point of the line AB is

The points with position vectors $60\hat i + 3\hat j,\,\,\,40\hat i - 8\hat j\,\,\,\,{\text{and}}\,\,\,a\hat i - 52\hat j$ are collinear, if $a$ is equal to

A vector of length 3 perpendicular to each of the vectors $3 i+j-4 k$ and $6 \mathbf{i}+5 \mathbf{j}-2 \mathbf{k}$ is

Let the unit vectors $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $A B C$. If $\vec{F}$ is the position vector of the mid point of the line segment joining its orthocentre and centroid then $(\vec{a}-\vec{F})^2+(\vec{b}-\vec{F})^2+($ $\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{F}})^2=$

The number of vectors of unit length perpendicular to the vectors $\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$

The position vector of the point where the line \vec r = \widehat i - \widehat j + \hat k + t( \widehat i + \widehat j - \hat k )$\mathbf{\vec r} = \widehat \mathbf{i} - \widehat \mathbf{j} + \mathbf{\hat k} + t\left( {\widehat \mathbf{i} + \widehat \mathbf{j} - \mathbf{\hat k}} \right)$ meets the plane \vec r·(\widehat i + \widehat j + \hat k) = 5$\mathbf{\vec r}\cdot(\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}) = 5$ is

Three vectors 7\widehat i - 11\widehat j + \hat k, 5\widehat i + 3\widehat j - 2\hat k$\widehat \mathbf{i} - 11\widehat \mathbf{j} + \mathbf{\hat k},\;5\widehat \mathbf{i} + 3\widehat \mathbf{j} - 2\mathbf{\hat k}$ and 12\widehat i - 8\widehat j - \hat k$12\widehat \mathbf{i} - 8\widehat \mathbf{j} - \mathbf{\hat k}$ from

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

The value of $a$ for which the lines $\frac{x-2}{1}=\frac{y-9}{2}=\frac{z-13}{3}$ and $\frac{x-a}{-1}=\frac{y-7}{2}=\frac{z+2}{-3}$ intersect, is