JEE Main 2002 — Vectors & 3D Geometry

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

If $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$ then $\vec{a}+\vec{b}+\vec{c}=$

$\vec{a}=3 \hat{i}-5 \hat{j}$ and $\vec{b}=6 \hat{i}+3 \hat{j}$ are two vectors and $\vec{c}$ is a vector such that $\vec{c}=\vec{a} \times \vec{b}$ then $$ |\vec{a}|:|\vec{b}|:|\vec{c}| $$

If $|a|=5,|b|=4,|c|=3$ thus what will be the value of $|a \cdot b+b . c+c . a|$, given that $\vec{a}+\vec{b}+\vec{c}=0$

If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\vec{a}|=7,|\vec{b}|=5,|\vec{c}|=3$ then angle between vector $\vec{b}$ and $\vec{c}$ is

If $|\vec{a}|=4,|\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi / 6$ then $(\vec{a} \times \vec{b})^2=2$ is equal to

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