Question

The value of \(\lambda\), for which the projection of \(\vec{a} = \lambda \hat{i} + \hat{j} +4 \hat{k}\) on \(\vec{b} =2\hat{i} +6 \hat{j} +3\hat{k}\) is 4 units

• A. 8
• B. 10
• C. 11
• D. 5

Answer 1

The Correct Option is D

To find the value of \(\lambda\) for which the projection of the vector \(\vec{a} = \lambda \hat{i} + \hat{j} + 4 \hat{k}\) on the vector \(\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\) is 4 units, we use the formula for the projection of \(\vec{a}\) onto \(\vec{b}\):
\[\text{Projection of }\vec{a}\text{ on }\vec{b}=\frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|}\]
First, calculate the dot product \(\vec{a}\cdot\vec{b}\):
\(\vec{a}\cdot\vec{b}=(\lambda \hat{i} + \hat{j} + 4\hat{k})\cdot(2\hat{i} + 6\hat{j} + 3\hat{k})=2\lambda + 6 + 12=2\lambda + 18\)
Next, calculate the magnitude \(\|\vec{b}\|\):
\(\|\vec{b}\|=\sqrt{2^2+6^2+3^2}=\sqrt{4+36+9}=\sqrt{49}=7\)
To equate the projection to 4 units, solve:
\(\frac{2\lambda+18}{7}=4\)
Multiply both sides by 7:
\(2\lambda+18=28\)
Subtract 18 from both sides:
\(2\lambda=10\)
Divide by 2:
\(\lambda=5\)
Thus, the value of \(\lambda\) is 5.