Question
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 , b=i−4j+5k and c=3i+j−k . Find the vector which is perpendicular to both a and b whose magnitude is twenty one times the magnitude of c.
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 ax bx jaybykazbz
p=∣p∣p
∣p∣=px2+py2+pz2
Complete step-by-step answer :
There are three vectors given in the question and that are a=4i+5j−k , b=i−4j+5k and c=3i+j−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×b . Let a perpendicular vector to a and b be vector p. Therefore, p=a×b .
Substitute the values of a and b.
⇒p=(4i+5j−k)×(i−4j+5k) .
The result of cross product is found by the determinant.
p=a×b=i ax bx jaybykazbz
⇒p=i 4 1 j5−4k−15=(5(5)−(−4)(−1))i−(4(5)−(1)(−1))j+(4(−4)−(5)(1))k
⇒p=(25−4)i−(20+1)j+(−16−5)k=21i−21j−21k .
Hence, we found a vector perpendicular vector to a and b.
The unit vector along a vector p is given as p=∣p∣p …. (i),
where p is the unit vector and ∣p∣ is the magnitude of the vector.
Magnitude of vector p will be ∣p∣=px2+py2+pz2 .
⇒∣p∣=212+(−21)2+(−21)2=3(21)2=213 .
Substitute the vector p and its magnitude in (i).
⇒p=21321i−21j−21k=3i−j−k .
Let the vector that we have to find be m.
Therefore, the unit vector along the vector is p=3i−j−k .
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
Therefore, the magnitude of vector m is ∣m∣=21∣c∣=2111 .
This means that the vector m is m=∣m∣p .
⇒m=2111(3i−j−k)=32111(i−j−k) .
Hence, we found the required vector.
So, the correct answer is “m=2111(3i−j−k)=32111(i−j−k) .”.
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.