The vector having magnitude equal to 3 and perpendicular to two vector

1.	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)
Answer» 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

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)

Answer»

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

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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)​

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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)