Exams Joint Entrance Examination Mathematics Vector Algebra Question 197 Question 197 of 379 Medium MCQ
Let a ⃗ , b ⃗ \vec a,\vec b a , b be vectors along the diagonals of a parallelogram of area 2 2 2\sqrt{2} 2 2 , angle between them acute, ∣ a ⃗ ∣ = 1 |\vec a|=1 ∣ a ∣ = 1 , and a ⃗ ⋅ b ⃗ = ∣ a ⃗ × b ⃗ ∣ \vec a\cdot\vec b=|\vec a\times\vec b| a ⋅ b = ∣ a × b ∣ . If c ⃗ = 2 2 ( a ⃗ × b ⃗ ) − 2 b ⃗ \vec c=2\sqrt{2}(\vec a\times\vec b)-2\vec b c = 2 2 ( a × b ) − 2 b , the angle between b ⃗ \vec b b and c ⃗ \vec c c is:
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Area = 1 2 ∣ a ⃗ × b ⃗ ∣ = 2 2 ⇒ ∣ a ⃗ × b ⃗ ∣ = 4 2 =\dfrac12|\vec a\times\vec b|=2\sqrt{2} \Rightarrow |\vec a\times\vec b|=4\sqrt{2} = 2 1 ∣ a × b ∣ = 2 2 ⇒ ∣ a × b ∣ = 4 2 .
a ⃗ ⋅ b ⃗ = ∣ a ⃗ × b ⃗ ∣ \vec a\cdot\vec b=|\vec a\times\vec b| a ⋅ b = ∣ a × b ∣ : ∣ b ⃗ ∣ cos θ = ∣ b ⃗ ∣ sin θ ⇒ tan θ = 1 ⇒ θ = π 4 |\vec b|\cos\theta = |\vec b|\sin\theta \Rightarrow \tan\theta=1 \Rightarrow \theta=\dfrac\pi4 ∣ b ∣ cos θ = ∣ b ∣ sin θ ⇒ tan θ = 1 ⇒ θ = 4 π (acute).
∣ b ⃗ ∣ sin θ = 4 2 ⇒ ∣ b ⃗ ∣ ⋅ 1 2 = 4 2 ⇒ ∣ b ⃗ ∣ = 8 |\vec b|\sin\theta = 4\sqrt{2} \Rightarrow |\vec b|\cdot\dfrac1{\sqrt{2}}=4\sqrt{2} \Rightarrow |\vec b|=8 ∣ b ∣ sin θ = 4 2 ⇒ ∣ b ∣ ⋅ 2 1 = 4 2 ⇒ ∣ b ∣ = 8 .
c ⃗ ⋅ b ⃗ = 2 2 ( a ⃗ × b ⃗ ) ⋅ b ⃗ − 2 ∣ b ⃗ ∣ 2 = 0 − 2 ( 64 ) = − 128 \vec c\cdot\vec b = 2\sqrt{2}(\vec a\times\vec b)\cdot\vec b - 2|\vec b|^2 = 0-2(64)=-128 c ⋅ b = 2 2 ( a × b ) ⋅ b − 2∣ b ∣ 2 = 0 − 2 ( 64 ) = − 128 .
∣ c ⃗ ∣ 2 = 8 ∣ a ⃗ × b ⃗ ∣ 2 + 4 ∣ b ⃗ ∣ 2 = 8 ( 32 ) + 4 ( 64 ) = 512 ⇒ ∣ c ⃗ ∣ = 16 2 |\vec c|^2 = 8|\vec a\times\vec b|^2+4|\vec b|^2 = 8(32)+4(64)=512 \Rightarrow |\vec c|=16\sqrt{2} ∣ c ∣ 2 = 8∣ a × b ∣ 2 + 4∣ b ∣ 2 = 8 ( 32 ) + 4 ( 64 ) = 512 ⇒ ∣ c ∣ = 16 2 .
\[\cos\alpha = \frac{-128}{8\times16\sqrt{2}} = -\frac1{\sqrt{2}} \implies \alpha=\frac{3\pi}4\] Answer the question to unlock