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Question

Let a=4^i+5^j^k,b=^i4^j+5^k and c=3^i+^j^k.
Find a vector d which is perpendicular to both a and b, and is such that dc=21.

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Solution

Given to us that,

a=4i+5jk

b=i4j+5k

c=3i+jk

Let d=xi+yj+zk

Therefore according to question:-

d.c=21

(xi+yj+zk).(3i+jk)=21(becausei.i=j.j=k.k=1)

On solving the above equation we get 3x+yz=21 ................(let equation 1)
We know that when two vectors are perpendicular then dot product of two vectors is equal to zero

Therefore d.a=0

(xi+yj+zk).(4i+5jk)=0

i.e. 4x+5yz=0...................(let equation 2)

Therefore d.b=0

(xi+yj+zk).(i4j+5k)=0

i.e. x4y+5z=0 ..................(let equation 3)

Now on adding equation1 and equation 3 and subtract it sum from equation 2
We get 8y5z=21.................(let equation 4)

On multiplying equation 3 by 3 and subtract it from equation 1
We get 13y16z=21................(let equation 5)

On solving equation 4 and equation 5

We get y=z=7

Now on putting value of y and z in equation equation 3

We get x=7

Therefore d=7i7j7k

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