Question

Let $ABCD$ be a parallelogram. If
$AB=\hat{i}+3\hat{j}+7\hat{k},AD=2\hat{i}+3\hat{j}-5\hat{k}$ and $p$ is a unit vector parallel to $AC$, then $p$ is equal to

• A. $\frac{1}{3}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$
• B. $\frac{1}{3}\left(2\hat{i}-2\hat{j}+\hat{k}\right)$
• C. $\frac{1}{7}\left(3\hat{i}+6\hat{j}+2\hat{k}\right)$
• D. $\frac{1}{7}\left(6\hat{i}+2\hat{j}+3\hat{k}\right)$
• E. $\frac{1}{7}\left(6\hat{i}+2\hat{j}-3\hat{k}\right)$

Answer 1

Answer: (C)

$AC=AB+AD$
$=(\hat{i}+3\hat{j}+7\hat{k})+(2\hat{i}+3\hat{j}-5\hat{k})$
$=3\hat{i}+6\hat{j}+2\hat{k}$

Now, $p$ is a unit vector parallel to AC
Then, $p=\frac{AC}{|AC|}$
$=\frac{3\hat{i}+6\hat{j}+2\hat{k}}{\sqrt{9+36+4}}$
$\Rightarrow p=\frac{1}{7}(3\hat{i}+6\hat{j}+2\hat{k})$