Notes : Projection of a Vector Along Another Vector - Class 11 Physics

PROJECTION OF A VECTOR ALONG ANOTHER VECTOR.

Consider two vectors \(\vec{A}\) and \(\vec{B}\)

Then the projection of vector \(\vec{A}\) in the direction of vector \(\vec{B}\) is given by

\(P = |\vec{A}| \cos \theta\)

Since \(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}\)

$\therefore P = \frac{|\vec{A}| (\vec{A} \cdot \vec{B})}{|\vec{A}| |\vec{B}|} = \vec{A} \cdot \frac{\vec{B}}{|\vec{B}|}$

\(( \because \vec{B} = |\vec{B}| \hat{B} )\)

$P= \vec{A} \cdot \hat{B}$

Thus, projection of a vector \(\vec{A}\) in the direction of a vector \(\vec{B}\) = Vector \(\vec{A}\) \(\cdot\) unit vector in the direction of \(\vec{B}\).

and projection of vector \(\vec{B}\) in the direction of vector \(\vec{A}\) is given by

\(P' = |\vec{B}| \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|} = \vec{B} \cdot \hat{A}\)

Thus, projection of vector \(\vec{B}\) in the direction of a vector \(\vec{A}\) = Vector \(\vec{B}\) \(\cdot\) unit vector in the direction of \(\vec{A}\).

Numerical Example : 

Find the projection of vector \(\vec{A}\) in the direction of vector \(\vec{B}\), if \(\vec{A} = \hat{i} - 2\hat{j} + 2\hat{k}\) and \(\vec{B} = 2\hat{i} + \hat{j} + \hat{k}\).

SOLUTION. Projection of vector \(\vec{A}\) in the direction of vector \(\vec{B}\)

\(P = \vec{A} \cdot \hat{B}\)

Here, \(\vec{A} = \hat{i} - 2\hat{j} + 2\hat{k}\), \(\vec{B} = 2\hat{i} + \hat{j} + \hat{k}\)

$\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{(2)^2 + (1)^2 + (1)^2}}$

 $\hat{B}= \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$

Using these values in eqn. (1), we get

$P = (\hat{i} - 2\hat{j} + 2\hat{k}) \cdot \frac{(2\hat{i} + \hat{j} + \hat{k})}{\sqrt{6}} = \frac{1}{\sqrt{6}}(2 - 2 + 2)$

$P = \frac{2}{\sqrt{6}}$