Question

If \[\overrightarrow x = 3\hat i - 6\hat j - \hat k\], \[\overrightarrow y = \hat i + 4\hat j - 3\hat k\] and \[\overrightarrow z = 3\hat i - 4\hat j - 12\hat k\], then the magnitude of the projection of \[\overrightarrow x \times \overrightarrow y \] on \[\overrightarrow z \] is
(A) 12
(B) 15
(C) 14
(D) 13

Answer 1

Hint: Here first we will find the vector \[\overrightarrow x \times \overrightarrow y \] using the cross product and then we will find the unit vector of z and then finally we will find the projection of \[\overrightarrow x \times \overrightarrow y \] on \[\overrightarrow z \].
The unit vector of vector \[\overrightarrow a \] is given by:-
\[\hat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}\]
The projection of vector \[\overrightarrow b \] on \[\overrightarrow a \] is given by:-
\[{\text{Projection}} = \overrightarrow b .\hat a\]

Complete step-by-step answer:
The vector x is given by:-
\[\overrightarrow x = 3\hat i - 6\hat j - \hat k\]
The vector y is given by:-
\[\overrightarrow y = \hat i + 4\hat j - 3\hat k\]
Now we will find the vector \[\overrightarrow x \times \overrightarrow y \]
The cross product of x and y is given by:-
\[\overrightarrow x \times \overrightarrow y = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  3&{ - 6}&{ - 1} \\
  1&4&{ - 3}
\end{array}} \right|\]
Expanding the determinant with respect to column 1 we get:-
\[\overrightarrow x \times \overrightarrow y = \hat i\left[ {\left( { - 6} \right)\left( { - 3} \right) - \left( 4 \right)\left( { - 1} \right)} \right] - \hat j\left[ {\left( 3 \right)\left( { - 3} \right) - \left( 1 \right)\left( { - 1} \right)} \right] + \hat k\left[ {\left( 3 \right)\left( 4 \right) - \left( 1 \right)\left( { - 6} \right)} \right]\]
Simplifying it further we get:-
\[
  \overrightarrow x \times \overrightarrow y = \hat i\left[ {18 + 4} \right] - \hat j\left[ { - 9 + 1} \right] + \hat k\left[ {12 + 6} \right] \\
   \Rightarrow \overrightarrow x \times \overrightarrow y = 22\hat i + 8\hat j + 18\hat k \\
 \]
Now we will find the unit vector of z.
First we will find the magnitude of z.
The magnitude of a vector \[\overrightarrow A = a\hat i + b\hat j + c\hat k\] is given by:-
\[\left| {\overrightarrow A } \right| = \sqrt {{a^2} + {b^2} + {c^2}} \]
Hence applying this formula the magnitude of z is:-
\[\left| {\overrightarrow z } \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 12} \right)}^2}} \]
Simplifying it further we get:-
\[
  \left| {\overrightarrow z } \right| = \sqrt {9 + 16 + 144} \\
   \Rightarrow \left| {\overrightarrow z } \right| = \sqrt {169} \\
   \Rightarrow \left| {\overrightarrow z } \right| = 13 \\
 \]
Now we know that:
The unit vector of vector \[\overrightarrow a \] is given by:-
\[\hat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}\]
Hence the unit vector of z is given by:-
\[\hat z = \dfrac{{\overrightarrow z }}{{\left| {\overrightarrow z } \right|}}\]
Putting in the respective values we get:-

\[\hat z = \dfrac{{3\hat i - 4\hat j - 12\hat k}}{{13}}\]
Now we will find the projection of \[\overrightarrow x \times \overrightarrow y \] on \[\overrightarrow z \]
Now we know that the projection of vector \[\overrightarrow b \] on \[\overrightarrow a \] is given by:-
\[{\text{Projection}} = \overrightarrow b .\hat a\]
Hence the projection of \[\overrightarrow x \times \overrightarrow y \] on \[\overrightarrow z \] is given by:-
\[{\text{Projection}} = \overrightarrow x \times \overrightarrow y .\hat z\]
Hence putting the respective values we get:-
\[{\text{Projection}} = \left( {22\hat i + 8\hat j + 18\hat k} \right).\dfrac{{3\hat i - 4\hat j - 12\hat k}}{{13}}\]
Solving it further we get:-
We know that:-
\[
  \hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1 \\
  \hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0 \\
 \]
Hence we get:-
\[
  {\text{Projection}} = \dfrac{{22\left( 3 \right) + 8\left( { - 4} \right) + 18\left( { - 12} \right)}}{{13}} \\
   \Rightarrow {\text{Projection}} = \dfrac{{66 - 32 - 216}}{{13}} \\
   \Rightarrow {\text{Projection}} = \dfrac{{ - 182}}{{13}} \\
 \]
Now since the magnitude is the modulus value.
Therefore the magnitude of projection is:-
\[
  \left| {{\text{Projection}}} \right| = \left| {\dfrac{{ - 182}}{{13}}} \right| \\
   \Rightarrow \left| {{\text{Projection}}} \right| = 14 \\
 \]
Therefore the magnitude is 14.

So, the correct answer is “Option C”.

Note: Students should note that the projection of vector \[\overrightarrow b \] on \[\overrightarrow a \] is given by:-
\[{\text{Projection}} = \overrightarrow b .\hat a\]
While the projection of vector \[\overrightarrow a \] on \[\overrightarrow b \] is given by:-
\[{\text{Projection}} = \overrightarrow a .\hat b\]
So, students should not make mistakes in finding the projection.