Question

When the projection of vector $a=\lambda i+j+4k$ on vector $b=2i+6j+3k$ is $4$units. The value of $\lambda$ is?

• A. $0$
• B. $10$
• C. $5$
• D. $3$

Answer 1

The correct option is C

$5$

The explanation for the correct option:

Step 1: Given that:

Projection of $\overrightarrow{a}$and $\overrightarrow{b}$ is $4$units.

Step 2: Formula used:

We know that projection of $\overrightarrow{a}.\overrightarrow{b}$ is given by, $\frac{\overrightarrow{a.}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$

The projection is given by $4$ units, therefore $4=$$\frac{\overrightarrow{a.}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}...\left(1\right)$

Step 3: Solving the equation:

$$\begin{gathered} \left|\overrightarrow{b}\right|=\sqrt{2^2+6^2+3^2} \\ \left|\overrightarrow{b}\right|=\sqrt{4+36+9}=\sqrt{49}=7 \end{gathered}$$

Substitute $\left|\overrightarrow{b}\right|$ in equation $\left(1\right)$

$4=\frac{2\lambda +18}{7}$

$\begin{array}{rcl}& \Rightarrow & 28=2\lambda +18\\ & \Rightarrow & 10=2\lambda \\ & \Rightarrow & \lambda =\frac{10}{2}\\ & \Rightarrow & \lambda =5\end{array}$

Hence, option (C) is the correct answer.