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Question

# Vectors →a, →b and →c are of the same length and when taken pairwise, they form equal angles. If →a=^i+^j and →b=^j+^k, then vector →c can be

A
^i+^k
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B
13^i43^j13^k
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C
13^i+43^j13^k
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D
^i+2^j+^k
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Solution

## The correct option is C −13^i+43^j−13^k Let →c=x^i+y^j+z^k Then |→a|=|→b|=|→c|⇒x2+y2+z2=2 It is given that the angles between the vectors taken in pairs are equal, say θ. Therefore, cos θ=→a.→b|→a||→b|=0+1+0√2√2=12 ⇒→a.→c|→a||→c|=12 and →b.→c|→b||→c|=12 ⇒x+y2=12 and y+z2=12 ⇒x+y=1 and y+z=1 ⇒y=1−x and z=1−y=1−(1−x)=x x2+y2+z2=2⇒x2+(1−x)2+x2=2 ⇒(3x+1)(x−1)=0 ⇒x=1,−13 Now, y=1−x ⇒y=0 for x=1 and y=43 for x=−13 Hence, →c=^i+^k or →c=−13^i+43^j−13^k

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