The position vector of a point P is 2i^{-}+j^{-}+3k^{-} and a=-\stackrel{-}{i}-2k^{-}, b^{-}=i^{-}+j^{-}+2k^{-} are two vectors which determine a plane \pi. The equation of a line through P normal to b^{-} and lying on the plane \pi is

Correct Answer is : −r=2−i+−j+3−k+λ(−−i+5−j−2−k)

r^{-}=2i^{-}+j^{-}+3k^{-}+\lambda(-i^{-}+5j^{-}-2k^{-})

r^{-}=2i^{-}+j^{-}+3k^{-}+\lambda(i^{-}+j^{-}+k^{-})

r^{-}=2i^{-}+j^{-}+3k^{-}+\lambda(-2i^{-}-j^{-}+3k^{-})

r^{-}=2i^{-}+j^{-}+3k^{-}+\lambda(-3i^{-}+4j^{-}-5k^{-})

TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 30 of 160

Marks: +1, -0

The position vector of a point P is 2

−

and a=−

−

−2

−

are two vectors which determine a plane π. The equation of a line through P normal to

−

and lying on the plane π is

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