BPSC 47th Exam Prelims 2005 Solved Paper

Question : 126 of 150

Marks: +1, -0

The value of K so that points (1, 2, 3), (K, 0, 4) and (–2, 4, 2) may be collinear is –

Solution:

From the question. The points are collinear, so

|\table 1 , 2 , 3 ; k , 0 , 4 ; -2 , 4 , 2|=0

⇒ 1(0-16)-2(2 k+8)+3(4 k)=0

⇒-16-4 k-16+12 k=0

⇒ 8 k=32 ⇒ k=4

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