Question

$\overrightarrow{p}=w\hat{i}+x\hat{j}$ and $\overrightarrow{q}=y\hat{i}+z\hat{j}$ are two vectors in the first quadrant such that $|\overrightarrow{p}|=2|\overrightarrow{q}|=2r,r>0$ and $\overrightarrow{p}\cdot \overrightarrow{q}=0$. If $\overrightarrow{a}=w\hat{i}+2y\hat{j}$ and $\overrightarrow{b}=\frac{x}{2}\hat{i}+z\hat{j}$, then

• A. $|\overrightarrow{a}|=r$
• B. $|\overrightarrow{b}|=r$
• C. $\overrightarrow{a}\cdot \overrightarrow{b}=0$
• D. $|\overrightarrow{a}\times \overrightarrow{b}|=2r$

Answer 1

Answer: (B), (C)

$\overrightarrow{a}\cdot \overrightarrow{b}=0$

Here,
$|\overrightarrow{p}|=\sqrt{w^2+x^2}=2r...\left(1\right)$,
$|\overrightarrow{q}|=\sqrt{y^2+z^2}=r...\left(2\right)$,
$\overrightarrow{p}\cdot \overrightarrow{q}=wy+xz=0...\left(3\right)$

Now,
$|\overrightarrow{a}|=\sqrt{w^2+4y^2}...\left(4\right)$
$w=\frac{-xz}{y}$ from eqn $\left(3\right)$
Putting this value in eqn $\left(1\right)$, we get
$|\overrightarrow{p}|=\frac{x}{y}r=2r$
$\Rightarrow x=2y$

Similarly, putting $x=-\frac{wy}{z}$ in eqn $\left(1\right)$, we get
$w=2z$
Putting the value of $w$ in eqn $\left(4\right)$,
$|\overrightarrow{a}|=2r$
Similarly, we get $|\overrightarrow{b}|=r$
$\overrightarrow{a}\cdot \overrightarrow{b}=\frac{wx}{2}+2yz=wy+xz=0\left[\text{From (3)}\right]$
$|\overrightarrow{a}\times \overrightarrow{b}|=|wz-xy|=|2z^2-2y^2|=2|z^2-y^2|\ne 2r$