Question

If \(\overrightarrow A=(2\hat i+3\hat j−\hat k)m\) and \(\overrightarrow B=(\hat i+2\hat j+2\hat k)m\). The magnitude of component of vector \(\overrightarrow A\) along vector \(\overrightarrow B\) will be ______\(m\).

Answer 1

Correct Answer: 2

To find the magnitude of the component of vector \(\overrightarrow A\) along vector \(\overrightarrow B\), we use the formula for the component of a vector \(\overrightarrow A\) on \(\overrightarrow B\):
\[\text{Component of }\overrightarrow A\text{ along }\overrightarrow B = \frac{\overrightarrow A \cdot \overrightarrow B}{|\overrightarrow B|}\]
The dot product \(\overrightarrow A \cdot \overrightarrow B\) is calculated by:
\[\overrightarrow A \cdot \overrightarrow B = (2\hat i + 3\hat j - \hat k) \cdot (\hat i + 2\hat j + 2\hat k)\]
This equals:
\[(2 \times 1) + (3 \times 2) + (-1 \times 2) = 2 + 6 - 2 = 6\]
Then, find the magnitude of \(\overrightarrow B\):
\[|\overrightarrow B| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\]
Substitute into the component formula:
\[\text{Component} = \frac{6}{3} = 2\]
Therefore, the magnitude of the component of vector \(\overrightarrow A\) along vector \(\overrightarrow B\) is \(2m\) which falls within the range (2,2).