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Question

Mathematics Question on Magnitude and Directions of a Vector

The direction cosines of the vector a=−2i+j−ka=-2i+j-k are ?

A

2√6,1√6,1√6\dfrac{2}{√6}, \dfrac{1}{√6},\dfrac{1}{√6}

B

−2√6,1√6,−1√6\dfrac{-2}{√6}, \dfrac{1}{√6},\dfrac{-1}{√6}

C

−2√6,−1√6,−1√6\dfrac{-2}{√6}, \dfrac{-1}{√6},\dfrac{-1}{√6}

D

−2√6,1√6,1√6\dfrac{-2}{√6}, \dfrac{1}{√6},\dfrac{1}{√6}

E

−2√6,−1√6,1√6\dfrac{-2}{√6}, \dfrac{-1}{√6},\dfrac{1}{√6}

Answer

−2√6,1√6,−1√6\dfrac{-2}{√6}, \dfrac{1}{√6},\dfrac{-1}{√6}

Explanation

Solution

Given that

a=−2i+j−ka=-2i+j-k is a vector and we need to find the direction of Cosine of this vector

So we can proceed as

step1

∣a∣=√((−2)2+12+(−1)2)|a|= √((-2)^{2}+1^{2}+(-1)^{2})

  $=$$ √6$

_step 2 _

Direction of cosine of a vector are

−2√6,1√6,−1√6\dfrac{-2}{√6}, \dfrac{1}{√6},\dfrac{-1}{√6} (Ans.)