If $A, B, C$ and $D$ are the points with position vectors $\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$, $2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}$, $2\hat{\mathbf{i}} - 3\hat{\mathbf{k}}$ and $3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$ respectively, then find the projection of $AB$ along $CD$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\sqrt{21}$ ##
Here, $OA = \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$, $OB = 2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 3\hat{\mathbf{k}}$, $OC = 2\hat{\mathbf{i}} - 3\hat{\mathbf{k}}$ and $OD = 3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + \hat{\mathbf{k}}$
$∴AB = OB - OA = (2 - 1)\hat{\mathbf{i}} + (-1 - 1)\hat{\mathbf{j}} + (3 + 1)\hat{\mathbf{k}}$
$= \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 4\hat{\mathbf{k}}$
and $CD = OD - OC = (3 - 2)\hat{\mathbf{i}} + (-2 - 0)\hat{\mathbf{j}} + (1 + 3)\hat{\mathbf{k}}$$
$= \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 4\hat{\mathbf{k}}$
So, the projection of $AB$ along $CD = AB \cdot \frac{CD}{|CD|}$
$= \frac{(\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 4\hat{\mathbf{k}}) \cdot (\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 4\hat{\mathbf{k}})}{\sqrt{1^2 + 2^2 + 4^2}}$
$= \frac{1 + 4 + 16}{\sqrt{21}} = \frac{21}{\sqrt{21}}$
$= \sqrt{21}$ units