# Finding the projection of a 3D vector along the direction of (i-j)

The question is:-

Find the component of vector $$\vec A = a_x\hat i + a_y\hat j + a_z\hat k$$ along the direction of $$(\hat i - \hat j)$$

Since both the vector should be in the same dimensions to do a projection, I removed the k component from the first vector and was left with:-

$$\vec B = a_x\hat i + a_y\hat j$$ After this I used the scalar projection formula.

Let x be the projection of the B vector on the (i-j) vector

$$x = \frac{(a_x\hat i + a_y\hat j)(\hat i-\hat j)}{|\hat i - \hat j|}$$ $$x = \frac{a_x\hat i(\hat i) - a_x\hat i(\hat j) + a_y\hat j(\hat i) - a_y\hat j(\hat j)}{\sqrt{1^2 + (-1)^2}}$$ $$x = \frac{0 + a_x\hat k - a_y\hat k - 0)}{\sqrt{2}}$$ $$|x| = \frac{a_x-a_y}{\sqrt{2}}$$

Now that happens to be the correct answer but is my method of solving this correct? Thanks in advance.

• Something is fishy here. The direction "in the direction of $\hat i- \hat j$" is $\hat i- \hat j$. This direction is not $\hat k$. – zoli Jul 2 '15 at 15:37

## 1 Answer

The method is not correct

The problem is the mistaken way of using the scalar product operation. The scalar product of two vectors $\vec v_1 = v_{x_1}\vec i+ v_{y_1}\vec j$ and $\vec v_2=v_{x_2}\vec i+ v_{y_2}\vec j+v_{z_2}\vec k$ is

$$(v_{x_1}\vec i+ v_{y_1}\vec j+v_{z_1}\vec k)\cdot (v_{x_2}\vec i+ v_{y_2}\vec j+v_{z_2}\vec k)=v_{x_1}v_{x_2}+v_{y_1}v_{y_2}+v_{z_1}v_{z_2}.$$

The unit direction vector in the direction of $\vec v_1$ and $\vec 2$ is $d_{u}=\frac{\vec i - \vec j}{\sqrt 2}.$

The length of the projection of $\vec A = a_x\hat i + a_y\hat j + a_z\hat k$ onto $d_{u}$ is

$$\left(\frac{\vec i - \vec j}{\sqrt 2}\right)\cdot \left( a_x\vec i + a_y\vec j + a_z\vec k\right)=\frac{a_x- a_y}{\sqrt 2}$$

which number is the component of the vector $\vec A$ in the direction of $\vec i-\vec j$.