Vector Algebra — Class 12 Maths Solution

As we know that projection of $\overrightarrow {\rm{a}}$ along $\overrightarrow {\rm{b}}$

is $\frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{|\vec b|}}$.

Here, $\overrightarrow {{\rm{OA}}} = \widehat {\rm{i}} + \widehat {\rm{j}} - \widehat {\rm{k}},$

$\overrightarrow {{\rm{OB}}} = 2\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}},$ $\overrightarrow {{\rm{OC}}}$

$= 2\widehat {\rm{i}} - 3\widehat {\rm{k}}$

and $\overrightarrow {{\rm{OD}}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + \widehat {\rm{k}}$

$\therefore$ $\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}} = (2 - 1)\widehat {\rm{i}} + ( - 1 - 1)\widehat {\rm{j}} + (3 + 1)\widehat {\rm{k}}$

$= \widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$

and $\overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{OD}}} - \overrightarrow {{\rm{OC}}} = (3 - 2)\widehat {\rm{i}} + ( - 2 - 0)\widehat {\rm{j}} + (1 + 3)\widehat {\rm{k}}$

$= \widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$

So, the projection of $\overrightarrow {{\rm{AB}}}$

along $\overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{AB}}} \cdot \frac{{\overrightarrow {{\rm{CD}}} }}{{|\overrightarrow {{\rm{CD}}} |}}$

$= \frac{{(\widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}}) \cdot (\widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}})}}{{\sqrt {{1^2} + {2^2} + {4^2}} }}$

$= \frac{{1 + 4 + 16}}{{\sqrt {21} }} = \frac{{21}}{{\sqrt {21} }}$
$= \sqrt {21}$ units