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Question Number 56461 by cesar.marval.larez@gmail.com last updated on 16/Mar/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19
	$u_3 ^→ =au_1 ^→ +bu_2 ^→ =a(2i)+b(j−3k) =i(2a)+j(b)+k(−3b) u_3 ^→ ⊥V^→ so u_3 ^→ .V^→ =0 {i(2a)+j(b)+k(−3b)}.{i+j+k}=0 2a+b−3b=0 2a−2b=0→a=b b)u_3 ^→ =i(2a)+j(b)+k(−3b) =i(2a)+j(a)+k(−3a) a=b=2 u_3 ^→ =4i+2j−6k value of (√(4^2 +2^2 +(−6)^2 )) =(√(56)) director cosine={(4/(√(56))),(2/(√(56))),((−6)/(√(56)))}$
	${\vec{u}}_{3} = a {\vec{u}}_{1} + b {\vec{u}}_{2}$ $= a (2 i) + b (j - 3 k)$ $= i (2 a) + j (b) + k (- 3 b)$ ${\vec{u}}_{3} ⊥ \vec{V}$ $s o {\vec{u}}_{3} . \vec{V} = 0$ ${i (2 a) + j (b) + k (- 3 b)} . {i + j + k} = 0$ $2 a + b - 3 b = 0$ $2 a - 2 b = 0 \to a = b$ $b) {\vec{u}}_{3} = i (2 a) + j (b) + k (- 3 b)$ $= i (2 a) + j (a) + k (- 3 a) a = b = 2$ ${\vec{u}}_{3} = 4 i + 2 j - 6 k$ $v a l u e o f \sqrt{4^{2} + 2^{2} + {(- 6)}^{2}} = \sqrt{56}$ $d i r e c t o r c o s i n e = {\frac{4}{\sqrt{56}}, \frac{2}{\sqrt{56}}, \frac{- 6}{\sqrt{56}}}$
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