Solveeit Logo
Login

Question

Question: If \[\overrightarrow x = 3\hat i - 6\hat j - \hat k\], \[\overrightarrow y = \hat i + 4\hat j - 3\ha...

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

Explanation

Solution

Here first we will find the vector x→×y→\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 x→×y→\overrightarrow x \times \overrightarrow y on z→\overrightarrow z .
The unit vector of vector a→\overrightarrow a is given by:-
a^=a→∣a→∣\hat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}
The projection of vector b→\overrightarrow b on a→\overrightarrow a is given by:-
Projection=b→.a^{\text{Projection}} = \overrightarrow b .\hat a

Complete step-by-step answer:
The vector x is given by:-
x→=3i^−6j^−k^\overrightarrow x = 3\hat i - 6\hat j - \hat k
The vector y is given by:-
y→=i^+4j^−3k^\overrightarrow y = \hat i + 4\hat j - 3\hat k
Now we will find the vector x→×y→\overrightarrow x \times \overrightarrow y
The cross product of x and y is given by:-

{\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 \\

Henceweget:−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}} \\

Nowsincethemagnitudeisthemodulusvalue.Thereforethemagnitudeofprojectionis:−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.