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.
Best 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 $\vec i-\vec j$.