Q72 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics
If $\theta$ is the angle between the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$ and $\cos \theta=\frac{2}{3}$ then$2(a+b+3)=$
- A$a^2+b^2$
- B$a^2$
- C$b^2$
- Correct answerD$a b$
Solution
Given vectors are $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$.
Take,
$\begin{aligned}<br />& \cos \theta=\frac{(2 \hat{i}-\hat{j}+2 \hat{k})(a \hat{i}+4 \hat{j}+b \hat{k})}{\sqrt{2^2+(-1)^2+2^2} \sqrt{a^2+4^2+b^2}} \\<br />& \frac{2}{3}=\frac{2 a-4+2 b}{\sqrt{4+1+4} \sqrt{a^2+b^2+16}} \\<br />& \frac{2}{3}=\frac{2(a-2+b)}{3 \sqrt{a^2+b^2+16}} \\<br />& \sqrt{a^2+b^2+16}=((a+b)-2)<br />\end{aligned}$
Take square both sides.
$\begin{aligned}<br />& a^2+b^2+16=(a+b)^2+4-4(a+b) \\<br />& a^2+b^2+16=a^2+b^2+2 a b+4-4 a-4 b \\<br />& 4 a+4 b+12=2 a b \\<br />& 2 a+2 b+6=a b \\<br />& 2(a+b+3)=a b<br />\end{aligned}$
Therefore, required value is $a b$.
Take,
$\begin{aligned}<br />& \cos \theta=\frac{(2 \hat{i}-\hat{j}+2 \hat{k})(a \hat{i}+4 \hat{j}+b \hat{k})}{\sqrt{2^2+(-1)^2+2^2} \sqrt{a^2+4^2+b^2}} \\<br />& \frac{2}{3}=\frac{2 a-4+2 b}{\sqrt{4+1+4} \sqrt{a^2+b^2+16}} \\<br />& \frac{2}{3}=\frac{2(a-2+b)}{3 \sqrt{a^2+b^2+16}} \\<br />& \sqrt{a^2+b^2+16}=((a+b)-2)<br />\end{aligned}$
Take square both sides.
$\begin{aligned}<br />& a^2+b^2+16=(a+b)^2+4-4(a+b) \\<br />& a^2+b^2+16=a^2+b^2+2 a b+4-4 a-4 b \\<br />& 4 a+4 b+12=2 a b \\<br />& 2 a+2 b+6=a b \\<br />& 2(a+b+3)=a b<br />\end{aligned}$
Therefore, required value is $a b$.
Hence, the correct answer is option: 'D'.