Question

Let $ a=4\widehat{i}+5\widehat{j}-\widehat{k} $ , $ b=\widehat{i}-4\widehat{j}+5\widehat{k} $ and $ c=3\widehat{i}+\widehat{j}-\widehat{k} $ . Find the vector which is perpendicular to both a and b whose magnitude is twenty one times the magnitude of c.

Answer 1

Hint: Find the unit vector along the line perpendicular to the two vectors a and b by finding the cross product of the vectors a and b. Once you find the unit vector, find the magnitude of the vector with help of the given data. Then you can find the required vector.
Formula used:
 $ a\times b=\left[ \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   {{a}_{x}} & {{a}_{y}} & {{a}_{z}} \\
   {{b}_{x}} & {{b}_{y}} & {{b}_{z}} \\
\end{matrix} \right] $
 $ \widehat{p}=\dfrac{p}{\left| p \right|} $
 $ \left| p \right|=\sqrt{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}} $

Complete step-by-step answer:
There are three vectors given in the question and that are $ a=4\widehat{i}+5\widehat{j}-\widehat{k} $ , $ b=\widehat{i}-4\widehat{j}+5\widehat{k} $ and $ c=3\widehat{i}+\widehat{j}-\widehat{k} $ . We supposed to find the vector which is perpendicular to both a and b whose magnitude is twenty one times the magnitude of c.
Therefore, let us first find the unit vector along the perpendicular vector to a and b. For this, we will find the cross product of the vector a and b. A cross product is an operation on two vectors, which yields a vector perpendicular to the two vectors.
The cross product of two vectors a and b is given as $ a\times b $ . Let a perpendicular vector to a and b be vector p. Therefore, $ p=a\times b $ .
Substitute the values of a and b.
 $ \Rightarrow p=\left( 4\widehat{i}+5\widehat{j}-\widehat{k} \right)\times \left( \widehat{i}-4\widehat{j}+5\widehat{k} \right) $ .
The result of cross product is found by the determinant.
 $ p=a\times b=\left[ \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   {{a}_{x}} & {{a}_{y}} & {{a}_{z}} \\
   {{b}_{x}} & {{b}_{y}} & {{b}_{z}} \\
\end{matrix} \right] $
 $ \Rightarrow p=\left[ \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   4 & 5 & -1 \\
   1 & -4 & 5 \\
\end{matrix} \right]=\left( 5(5)-(-4)(-1) \right)\widehat{i}-\left( 4(5)-(1)(-1) \right)\widehat{j}+\left( 4(-4)-(5)(1) \right)\widehat{k} $
 $ \Rightarrow p=(25-4)\widehat{i}-(20+1)\widehat{j}+(-16-5)\widehat{k}=21\widehat{i}-21\widehat{j}-21\widehat{k} $ .
Hence, we found a vector perpendicular vector to a and b.
 The unit vector along a vector p is given as $ \widehat{p}=\dfrac{p}{\left| p \right|} $ …. (i),
where $ \widehat{p} $ is the unit vector and $ \left| p \right| $ is the magnitude of the vector.
Magnitude of vector p will be $ \left| p \right|=\sqrt{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}} $ .
 $ \Rightarrow \left| p \right|=\sqrt{{{21}^{2}}+{{(-21)}^{2}}+{{(-21)}^{2}}}=\sqrt{3{{(21)}^{2}}}=21\sqrt{3} $ .
Substitute the vector p and its magnitude in (i).
 $ \Rightarrow \widehat{p}=\dfrac{21\widehat{i}-21\widehat{j}-21\widehat{k}}{21\sqrt{3}}=\dfrac{\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{3}} $ .
Let the vector that we have to find be m.
Therefore, the unit vector along the vector is $ \widehat{p}=\dfrac{\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{3}} $ .
It is given that the magnitude of vector m is 21 times the magnitude of vector c.
And the magnitude of vector c is $ \left| c \right|=\sqrt{{{3}^{2}}+{{(1)}^{2}}+{{(-1)}^{2}}}=\sqrt{9+1+1}=\sqrt{11} $
Therefore, the magnitude of vector m is $ \left| m \right|=21\left| c \right|=21\sqrt{11} $ .
This means that the vector m is $ m=\left| m \right|\widehat{p} $ .
 $ \Rightarrow m=21\sqrt{11}\left( \dfrac{\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{3}} \right)=\dfrac{21\sqrt{11}}{\sqrt{3}}\left( \widehat{i}-\widehat{j}-\widehat{k} \right) $ .
Hence, we found the required vector.
So, the correct answer is “$m=21\sqrt{11}\left( \dfrac{\widehat{i}-\widehat{j}-\widehat{k}}{\sqrt{3}} \right)=\dfrac{21\sqrt{11}}{\sqrt{3}}\left( \widehat{i}-\widehat{j}-\widehat{k} \right) $ .”.

Note: Note that cross product is vector operation and the result of the cross product is also a vector. Hence, it is also called a vector product. We can perform the operation of cross products with scalar quantities.