Vectors & 3D Geometry PYQ — Page 20

With two forces acting at a point, the maximum effect is obtained when their resultant is $4 \mathrm{~N}$. If they act at right angles, then their resultant is $3 \mathrm{~N}$. Then the forces are

Three forces $\vec{P}, \vec{Q}$ and $\vec{R}$ acting along IA, IB and IC, where I is the incentre of a $\triangle A B C$, are in equilibrium. Then $\vec{P}: \vec{Q}: \vec{R}$ is

Let $\bar{u}, \bar{v}, \bar{w}$ be such that $|\bar{u}|=1,|\bar{v}|=2,|\bar{w}|=3$. If the projection $\bar{v}$ along $\bar{u}$ is equal to that of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|$ equals

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

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be non-zero vectors such that $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the acute angle between the vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$, then $\sin \theta$ equals

In a right angle $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}$ and sides a, b, c are respectively, $5 \mathrm{~cm}, 4 \mathrm{~cm}$ and $3 \mathrm{~cm}$. If a force $\vec{F}$ has moments 0,9 and 16 in $N$ cm. units respectively about vertices $A, B$ and $C$, then magnitude of $\vec{F}$ is

If the straight lines $x=1+s, y=-3-\lambda s, z=1+\lambda s$ and $x=\frac{t}{2}, y=1+t, z=2-t$ with parameters $s$ and $t$ respectively, are co-planar then $\lambda$ equals

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}, \lambda \overline{\mathrm{b}}+4 \overline{\mathrm{c}}$ and $(2 \lambda-1) \overline{\mathrm{c}}$ are non-coplanar for

A velocity $\frac{1}{4} \mathrm{~m} / \mathrm{s}$ is resolved into two components along $\mathrm{OA}$ and $\mathrm{OB}$ making angles $30^{\circ}$ and $45^{\circ}$ respectively with the given velocity. Then the component along $\mathrm{OB}$ is

A particle is acted upon by constant forces $4 I+J-3 k$ and $3 I+J-k$ which displace it from a point $\hat{i}+2 \hat{j}+3 \hat{k}$ to the point $5 \hat{i}+4 \hat{j}+\hat{k}$. The work done in standard units by the forces is given by

A line with direction cosines proportional to $2,1,2$ meets each of the lines $x=y+a=z$ and $x+a=2 y=2 z$. The co-ordinates of each of the point of intersection are given by

The two lines $x=a y+b, z=c y+d$ and $x=a^{\prime} y+b^{\prime} z=c^{\prime} y+d^{\prime}$ will be perpendicular, if and only if

The resultant of forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$. If $\overrightarrow{\mathrm{Q}}$ is doubled then $\overrightarrow{\mathrm{R}}$ is doubled. If the direction of $\overrightarrow{\mathrm{Q}}$ is reversed, then $\overrightarrow{\mathrm{R}}$ is again doubled. Then $\mathrm{P}^2: \mathrm{Q}^2: \mathrm{R}^2$ is

The lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-\mathrm{k}}$ and $\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}$ are coplanar if

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

If $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ and vectors $\left(1, a, a^2\right),\left(a, b, b^2\right)$ and $\left(a, c, c^2\right)$ are non-coplanar, then the product abc equals

If $\vec{u}, \vec{v}$ and $\vec{w}$ are three non-coplanar vectors, then $(\vec{u}+\vec{v}-\vec{w}) .(\vec{u}-\vec{v}) \times(\vec{v}-\vec{w})$ equals

Consider points A, B, C and D with position vectors $7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ respectively. Then $\mathrm{ABCD}$ is a

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

A particle acted on by constant forces $4 \hat{i}+\hat{j}-3 \hat{k}$ and $3 \hat{i}+\hat{j}-\hat{k}$ to the point $5 \hat{i}+4 \hat{j}-\hat{k}$. The total work done by the forces is

A couple is of moment $\overrightarrow{\mathrm{G}}$ and the force forming the couple is $\overrightarrow{\mathrm{P}}$. If $\overrightarrow{\mathrm{P}}$ is turned through a right angle the moment of the couple thus formed is $\overrightarrow{\mathrm{H}}$. If instead, the force $\overrightarrow{\mathrm{P}}$ are turned through an angle $\alpha$, then the moment of couple becomes

$3 \lambda \vec{c}+2 \mu(\vec{a} \times \vec{b})=0$ then

The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are

The d.r. of normal to the plane through $(1,0,0),(0,1,0)$ which makes an angle $\pi / 4$ with plane $x+y=3$ are