consider the value of two vectors L=1i+2j+3k and l=4i+5j+6k.Find the value of the scalar A such that the vector L-Al is perpendicular to the L
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Given:
L=1i+2j+3k
I=4i+5j+6k
To Find :
Find a scalar A such that L-AI is perpendicular to L.
(L-AI).L=0
Explanation:
AI=A(4i+5j+6k)
L-AI= 1i+2j+3k - A(4i+5j+6k)
=(1-4A)i + (2-5A)j + (3-6A)k
(L-AI).L={(1-4A)i + (2-5A)j + (3-6A)k}.(1i+2j+3k)
=(1-4A)x1 + (2-5A)x2 + (3-6A)x3 (i.i=1.j.j=1,k.k=1)
=1-4A + 4 -10A +9- 18A
=14-32A
(L-AI).L =0
∴14-32A = 0
A= 7/16
Hence the value of the scalar A is 7/16.
Answered by
3
Answer:
Value of scalar is.
Explanation:
Given:
A is a scalar.
To find:
The value of scalar is.
Calculation:
As we know that if two vectors are perpendicular to each other that their dot product is equal to zero.
Hence, value of Scalar is
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