If \(\vec{a} = 4\hat{j}\) and \(\vec{b} = 3\hat{j} + 4\hat{k}\), then the vector form of the component of \(\vec{a}\) along \(\vec{b}\) is

Concept:

The vector form of the component of \(\vec{a}\) along \(\vec{b}\) = \(\rm \left(\dfrac {\vec a. \vec b}{|\vec b|^2}\right)\vec b\)

Calculations:

Given,  \(\vec{a} = 4\hat{j}\) and \(\vec{b} = 3\hat{j} + 4\hat{k}\)

⇒ \(\rm \vec a . \vec b = (4\vec j).(3\vec j + 4 \vec k)\)

⇒ \(\rm \vec a . \vec b = 12\)

and  \(\rm |\vec b| = 5 \)

The vector form of the component of \(\vec{a}\) along \(\vec{b}\) = \(\rm \left(\dfrac {\vec a. \vec b}{|\vec b|^2}\right)\vec b\)

= \(\rm \dfrac {12}{25}(3\vec j + 4 \vec k)\)

Hence, if \(\vec{a} = 4\hat{j}\) and \(\vec{b} = 3\hat{j} + 4\hat{k}\), then the vector form of the component of \(\vec{a}\) along \(\vec{b}\) is \(\rm \dfrac {12}{25}(3\vec j + 4 \vec k)\)