The vector having magnitude equal to 3 and perpendicular to two vectors A=2i+2j+k and B - 2i – 2j+3k is
(A) +(21-j-2k)
(B) +(3i+j-2k)
(C) -(3u–3–3k)
(D) (3i-j- 3k)
Answers
Answered by
94
Answer :
- Here we could actually use a shortcut and easily arrive at the final answer. If we have a look at options & if we just find the one whose magnitude is equal to 3, then it would be the answer.
So, for Option A :
➞ Magnitude of a vector = √(a²+b²+c²...)
➞ A = √(2²+1²+2²)
➞ A = √(4+1+4)
➞ A = √9
➞ A = 3
- Similarly if you calculate for the remaining options we would find that the magnitudes of those ar not equal to 3.
- So we may directly conclude that the final answer is Option A
Answered by
70
Given :-
- The vector having magnitude equal to 3 and perpendicular to two vectors A = 2î + 2ĵ + k^ and B = 2î - 2ĵ + 3k^.
To Find :-
- Vector?
Solution :-
- Option (A)
Given vector: 2î + ĵ + 2k^
Finding its magnitude,
➡ Magnitude = √(2² + 1² + 2²)
➡ Magnitude = √(2 × 2 + 1 × 1 + 2 × 2)
➡ Magnitude = √(4 + 1 + 4)
➡ Magnitude = √(8 + 1)
➡ Magnitude = √(9)
➡ Magnitude = √(3 × 3)
➡ Magnitude = 3
- Hence, vector 2î + ĵ + 2k^ have magnitude 3. So, required vector is option (A) 2î + ĵ + 2k^.
Learn more :-
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