If a-b-c are non-coplanar unit vectors such that a - b - c - b - c -2- then the angle between a and b isA- -4B- -2C- 3 -4D- -

The correct option is C 3π/4a× (b× c)=b+c/√(2)⇒ (a.c)b (a.b)c=b+c/√(2) ⇒ [ (a.c) 1/√(2) ]b [ (a.b)+1/√(2) ]c=0 ⇒ a.c=1/√(2),a.b= 1/√(2) ⇒ |a||c|cosθ = 1/√(2), ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

If a,b,c are ...

Question

If a,b,c are non-coplanar unit vectors such that $a \times (b \times c) = \frac{b + c}{\sqrt{2}}$ , then the angle between a and b is

\frac{π}{4}

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\frac{π}{2}

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\frac{3 π}{4}

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π

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Solution

The correct option is C $\frac{3 π}{4}$
$a \times (b \times c) = \frac{b + c}{\sqrt{2}} \Rightarrow (a . c) b - (a . b) c = \frac{b + c}{\sqrt{2}} \Rightarrow [(a . c) - \frac{1}{\sqrt{2}}] b - [(a . b) + \frac{1}{\sqrt{2}}] c = 0 \Rightarrow a . c = \frac{1}{\sqrt{2}}, a . b = - \frac{1}{\sqrt{2}}$
$\Rightarrow | a | | c | c o s θ = \frac{1}{\sqrt{2}}, | a | | b | c o s ϕ = - \frac{1}{\sqrt{2}} \Rightarrow c o s θ = \frac{1}{\sqrt{2}}, c o s = - \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}, ϕ = \frac{3 π}{4} .$

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If a,b,c are non-coplanar unit vectors such that a×(b×c)=b+c√2, then the angle between a and b is

The correct option is C 3π4a×(b×c)=b+c√2⇒(a.c)b−(a.b)c=b+c√2⇒[(a.c)−1√2]b−[(a.b)+1√2]c=0⇒a.c=1√2,a.b=−1√2 ⇒|a||c|cosθ=1√2,|a||b|cosϕ=−1√2⇒cosθ=1√2,cos=−1√2⇒θ=π4,ϕ=3π4.

If a,b,c are non-coplanar unit vectors such that $a \times (b \times c) = \frac{b + c}{\sqrt{2}}$ , then the angle between a and b is