JEE Main 2004 — Vectors & 3D Geometry

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

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 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

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

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

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

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

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 $\bar{u}, \bar{v}, \bar{w}$ 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

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

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