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Question

If x,y,z are units vectors such that x+y+z=a , (x×y)×z=b, ax=32,ay=74 and |a|=2, then which of the following is FALSE?

A
y is perpendicular to z
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B
y is perpendicular to b
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C
Angle between x and y is acute.
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D
Angle between x and zis obtuse.
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Solution

The correct option is B y is perpendicular to b
Given :x+y+z=a
|x+y+z|=|a||x+y+z|=2|x+y+z|2=4(x+y+z)(x+y+z)=43+2(xy+yz+zx)=4xy+yz+zx=12(1)

Now,
x+y+z=a
Taking dot product with x and y, we get
1+xy+xz=32xy+xz=12(2)xy+1+yz=74xy+yz=34(3)
From equations (1),(2) and (3), we get
yz=0xz=14xy=34
Therefore,
(x×y)×z=bz×(x×y)=b(zy)x(zx)y=b14y=b
Hence, b is parallel to y.

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