Question

If the vector $a=2i+3j+6k$ and $b$ are collinear and $\left|b\right|=21$, then $b$ is equal to:

• A. $\pm \left(2i+3j+6k\right)$
• B. $\pm 3\left(2i+3j+6k\right)$
• C. $\left(i+j+k\right)$
• D. $\pm 21\left(2i+3j+6k\right)$

Answer 1

The correct option is B

$\pm 3\left(2i+3j+6k\right)$

Explanation for the correct option :

Step 1. Find the vector $b$.

As the vector $a$ and $b$ are collinear so $a=kb$ where $k$ is a constant. And so $\left|a\right|=\left|k\right|\left|b\right|$.

Now for the vector $a=2i+3j+6k$ the magnitude is given as:

$\begin{array}{rcl}\left|a\right|& =& \sqrt{2^2+3^2+6^2}\\ & =& \sqrt{4+9+36}\\ & =& \sqrt{49}\\ & =& 7\end{array}$

Now in the equation $\left|a\right|=\left|k\right|\left|b\right|$ substitute $7$ for $\left|a\right|$ and $21$ for $\left|b\right|$.

$\begin{array}{rcl}7& =& \left|k\right|·21\\ \frac{7}{21}& =& \left|k\right|\\ \pm \frac{1}{3}& =& k\end{array}$

Now substitute the value of vector $a$ and $k$ in the equation $a=kb$.

$\begin{array}{rcl}a& =& kb\\ (2i+3j+6k)& =& \pm \frac{1}{3}b\\ \pm 3(2i+3j+6k)& =& b\end{array}$

Hence, the correct option is B.