Question

What is the unit vector parallel to $a=3i+4j-2k?$.
What vector should be added to a so that resultant is the unit vector $\overrightarrow{i}$?

• A. $\frac{1}{\sqrt{\left(29\right)}}(-3i-4j+2k)b=+2i-4j-2k.$
• B. $\frac{1}{\sqrt{\left(29\right)}}(3i+4j+2k),b=-2i+4j+2k.$
• C. $\frac{1}{\sqrt{\left(29\right)}}(3i+4j-2k),b=-2i-4j+2k.$
• D. $\frac{1}{\sqrt{\left(29\right)}}(-3i-4j+2k),b=2i+4j+2k.$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{\left(29\right)}}(3i+4j-2k),b=-2i-4j+2k.$

We have a $a=3i+4j-2k$
$\therefore \left|a\right|=\sqrt{9+16+4}=\sqrt{29}$
$\therefore$ the unit vector parallel to $a$
$=\frac{a}{\left|a\right|}=\frac{1}{\sqrt{\left(29\right)}}(3i+4j-2k).$
Now suppose $b$ is the vector which when added to $a$ gives the resultant $i.$
Then $a+b=i$
or $b=i-a=i-(3i+4j-2k)$
$\therefore b=-2i-4j+2k.$