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Given vecA + vecB + vecC + vecD = vec0, ...

Given `vecA + vecB + vecC + vecD = vec0`, which of the following statements is not correct ?

A

`vecA, vecB, vecC and vecD` must each be a null vector.

B

The magnitude of `(vecA + vecC)` equals the magnitude of `(vecB + vecD)`.

C

The magnitude of `vecA` can never be greater than the sum of the magnitudes of `vecB, vecC and vecD.`

D

`vecB + vecC` must lie in the plane of `vecA and vecD` if `vecA and vecD` are not collinear and in the line of `vecA and vecD`, if they are collinear.

Text Solution

Verified by Experts

The correct Answer is:
A
The statement is not correct. It is because `vecA + vecB + vecC + vecD` can be zero in many ways other than `vecA, vecB, vecC and vecD` being each a rull vector.
(b) The statement is correct as proved below
`vecA + vecB + vecC + vecD = vec0 or vecA + vecC = - (vecB + vecD)`
`therefore |vecA + vecC| = |vecB + vecD|`
(c ) The statement is correct and can be proved using triangular inequality.
`|vecB| + |vecC| ge|vecR|" "...(i)`
`|vecR| + |vecD| ge|vecA|" "...(ii)`
Using (i) and (ii), `|vecA| le |vecB| + |vecC| + |vecD|`
(d) THe statement is correct as proved below
`vecA + vecB + vecC + vecD = vec0 or vecA + (vecB + vecC) + vecD = vec0`
The resultant sum of three vectors `vecA, (vecB+vecC)` and `vecD` can be zero only if `(vecB + vecC)` lies in the plane of `vecA and vecD` and these three vectors are represented by the three sides of a triangle taken in one order. IF `vecA and vecD` are collinear, then `(vecB + vecC)` must be in the line of `vecA and vecD`, only then the vector sum of all the vectors will be zero.
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