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Question: Let \( a=4\widehat{i}+5\widehat{j}-\widehat{k} \) , \( b=\widehat{i}-4\widehat{j}+5\widehat{k} \) an...

Let a=4i^+5j^k^a=4\widehat{i}+5\widehat{j}-\widehat{k} , b=i^4j^+5k^b=\widehat{i}-4\widehat{j}+5\widehat{k} and c=3i^+j^k^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.

Explanation

Solution

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×b=[i^j^k^ axayaz bxbybz ]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]
p^=pp\widehat{p}=\dfrac{p}{\left| p \right|}
p=px2+py2+pz2\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=4i^+5j^k^a=4\widehat{i}+5\widehat{j}-\widehat{k} , b=i^4j^+5k^b=\widehat{i}-4\widehat{j}+5\widehat{k} and c=3i^+j^k^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×ba\times b . Let a perpendicular vector to a and b be vector p. Therefore, p=a×bp=a\times b .
Substitute the values of a and b.
p=(4i^+5j^k^)×(i^4j^+5k^)\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×b=[i^j^k^ axayaz bxbybz ]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]
p=[i^j^k^ 451 145 ]=(5(5)(4)(1))i^(4(5)(1)(1))j^+(4(4)(5)(1))k^\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}
p=(254)i^(20+1)j^+(165)k^=21i^21j^21k^\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 p^=pp\widehat{p}=\dfrac{p}{\left| p \right|} …. (i),
where p^\widehat{p} is the unit vector and p\left| p \right| is the magnitude of the vector.
Magnitude of vector p will be p=px2+py2+pz2\left| p \right|=\sqrt{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}} .
p=212+(21)2+(21)2=3(21)2=213\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).
p^=21i^21j^21k^213=i^j^k^3\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 p^=i^j^k^3\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 c=32+(1)2+(1)2=9+1+1=11\left| c \right|=\sqrt{{{3}^{2}}+{{(1)}^{2}}+{{(-1)}^{2}}}=\sqrt{9+1+1}=\sqrt{11}
Therefore, the magnitude of vector m is m=21c=2111\left| m \right|=21\left| c \right|=21\sqrt{11} .
This means that the vector m is m=mp^m=\left| m \right|\widehat{p} .
m=2111(i^j^k^3)=21113(i^j^k^)\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=2111(i^j^k^3)=21113(i^j^k^)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.