# Thread: Find the projection of a vector

1. ## Find the projection of a vector

Find the projection of the vector $
5 \hat{i} + 2 \hat{j} + 3 \hat{k}$
along $2\hat{i} + \hat{j} + \hat{k}.
$

2. Originally Posted by zorro
Find the projection of the vector $
5 \hat{i} + 2 \hat{j} + 3 \hat{k}$
along $2\hat{i} + \hat{j} + \hat{k}.
$
Hi zorro.

Let $\mathbf u=5\mathbf i+2\mathbf j+3\mathbf k$ and $\mathbf v=2\mathbf i+\mathbf j+\mathbf k.$

You want to find a vector parallel to $\mathbf v$ of magnitude $|\mathbf u|\cos\theta$ where $\theta$ is the angle between $\mathbf u$ and $\mathbf v.$ This would be the vector $\frac{|\mathbf u|\cos\theta}{|\mathbf v|}\mathbf v.$

Since $\cos\theta\ =\ \frac{\mathbf{u\cdot v}}{|\mathbf u||\mathbf v|}$ the required vector is $\frac{\mathbf{u\cdot v}}{|\mathbf v|^2}\mathbf v=\frac{5\cdot2+2\cdot1+3\cdot1}{\left(\sqrt{2^2+1 ^2+1^2}\right)^2}\left(2\mathbf i+\mathbf j+\mathbf k\right)=\frac{15}{\sqrt6}\left(2\mathbf i+\mathbf j+\mathbf k\right).$

3. ## My work?

Originally Posted by TheAbstractionist
Hi zorro.

Let $\mathbf u=5\mathbf i+2\mathbf j+3\mathbf k$ and $\mathbf v=2\mathbf i+\mathbf j+\mathbf k.$

You want to find a vector parallel to $\mathbf v$ of magnitude $|\mathbf u|\cos\theta$ where $\theta$ is the angle between $\mathbf u$ and $\mathbf v.$ This would be the vector $\frac{|\mathbf u|\cos\theta}{|\mathbf v|}\mathbf v.$

Since $\cos\theta\ =\ \frac{\mathbf{u\cdot v}}{|\mathbf u||\mathbf v|}$ the required vector is $\frac{\mathbf{u\cdot v}}{|\mathbf v|^2}\mathbf v=\frac{5\cdot2+2\cdot1+3\cdot1}{\left(\sqrt{2^2+1 ^2+1^2}\right)^2}\left(2\mathbf i+\mathbf j+\mathbf k\right)=\frac{15}{\sqrt6}\left(2\mathbf i+\mathbf j+\mathbf k\right).$

Projection of the vector = $5i + 2j + 3k \times \frac{2i + j + k}{\sqrt{4+1+1}}$

$= \frac{15}{\sqrt{6}}$..........Is this method correct and is the result also correct ?

4. Originally Posted by zorro
Projection of the vector = $5i + 2j + 3k \times \frac{2i + j + k}{\sqrt{4+1+1}}$

$= \frac{15}{\sqrt{6}}$..........Is this method correct and is the result also correct ?

You use $\times$ Do you mean using cross product would also reach the answer ? No ... cross product or vector product gives vector not scalar $\frac{15}{\sqrt{6}}$ ...

5. Originally Posted by simplependulum
You use $\times$ Do you mean using cross product would also reach the answer ? No ... cross product or vector product gives vector not scalar $\frac{15}{\sqrt{6}}$ ...
$\times$ is not a cross product here it just means a multiplication sign .......i am sorry for coding that there