For what value of m the vector a is 2i^+3j^+6k^ is perpendicular to vector b 3i^-mj^-6k^
Answers
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Answer :
m = - 10
Explanation :
We are given with two vectors
A = 2 i + 3 j + 6 k
and
B = 3 i - m j - 6 k
( where i , j , k are orthogonal vectors )
Since , they are perpendicular to each other
therefore, angle between them is θ = 90°
so,Dot product of these vectors will be
→ A . B = A B cos θ
→ A . B = A B cos 90°
[∵ cos 90° = 0 , ∴ ]
→ A . B = 0
→ ( 2 i + 3 j + 6 k ) . ( 3 i - m j - 6 k ) = 0
→ 2i 3i + 2i (-mj) + 2i (-6k) + 3j 3i + 3j (-mj) + 3j (-6k) + 6k 3i + 6k (-mj) + 6k (-6k) = 0
[ ∵ Dot product of an orthogonal vector with itself is always 1 and product of two different orthogonal vectors is always zero ]
→ 6 (1) -2 m (0) - 12 (0) + 9 (0) - 3 m (1) - 18 (0) + 18 (0) - 6 m (0) - 36 (1) = 0
→ 6 - 3 m - 36 = 0
→ - 3 m = 30
→ m = 30 / -3
→ m = -10
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The Dot product of two orthogonal vectors
of the form
ai + bj + ck
&
pi + qj + rk
is given by
=> (ai+bj+ck).(pi+qj+rk) = ap + bq + cr
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