Exams Joint Entrance Examination Mathematics Vector Algebra Question 49 Question 49 of 379 Medium MCQ
Let a → = 6 i ^ + j ^ − k ^ \overset{\to }{a}=6\hat{i}+\hat{j}-\hat{k} a → = 6 i ^ + j ^ − k ^ and b → = i ^ + j ^ \overset{\to }{b}=\hat{i}+\hat{j} b → = i ^ + j ^ . If c → \overset{\to }{c} c → is a is vector such that ∣ c → ∣ ≥ 6 , a → ⋅ c → = 6 ∣ c → ∣ , ∣ c → − a → ∣ = 2 2 |\overset{\to }{c}|\ge 6,\overset{\to }{a}\cdot \overset{\to }{c}=6|\overset{\to }{c}|,|\overset{\to }{c}-\overset{\to }{a}|=2\sqrt{2} ∣ c → ∣ ≥ 6 , a → ⋅ c → = 6∣ c → ∣ , ∣ c → − a → ∣ = 2 2 and the angle between a ⃗ × b ⃗ \vec{a}\times \vec{b} a × b and c ⃗ \vec{c} c is 60 ∘ {60}^{\circ } 60 ∘ , then ∣ ( a ⃗ × b ⃗ ) × c ⃗ ∣ |(\vec{a}\times \vec{b})\times \vec{c}| ∣ ( a × b ) × c ∣ is equal to:
Solutions · 1 Post your solution Official solution
∣ ( a ⃗ × b ⃗ ) × c ⃗ ∣ = ∣ a ⃗ × b ⃗ ∣ ∣ c ⃗ ∣ sin 60 ∘ ∣ i j k 6 1 − 1 1 1 0 ∣ = i ( 1 ) − j ( 1 ) + k ( 5 ) = i − j + 5 k ∣ a ⃗ × b ⃗ ∣ = 1 + 1 + 25 = 27 ∣ c ⃗ − a ⃗ ∣ = 2 2 c 2 + a 2 − 2 a c = 8 c 2 − 12 c + 30 = 0 c = 12 + 24 2 = 6 + 6 ⇒ ∣ ( a ⃗ × b ⃗ ) × c ⃗ ∣ = 27 × ( 6 + 6 ) × 3 2 = a 2 ( 6 + 6 ) \begin{gathered} |(\vec{a}\times \vec{b})\times \vec{c}|=|\vec{a}\times \vec{b}||\vec{c}|\sin {60}^{\circ } \\ |\begin{array}{|c|}i \quad j \quad k \\ 6 \quad 1 \quad -1 \\ 1 \quad 1 \quad 0\end{array}|=i(1)-j(1)+k(5) \\ =i-j+5k \\ |\vec{a}\times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\ |\vec{c}-\vec{a}|=2\sqrt{2} \\ {c}^{2}+{a}^{2}-2ac=8 \\ {c}^{2}-12c+30=0 \\ c=\frac{12+\sqrt{24}}{2}=6+\sqrt{6} \\ \Rightarrow |(\vec{a}\times \vec{b})\times \vec{c}|=\sqrt{27}\times (6+\sqrt{6})\times \frac{\sqrt{3}}{2} \\ =\frac{a}{2}(6+\sqrt{6}) \end{gathered} ∣ ( a × b ) × c ∣ = ∣ a × b ∣∣ c ∣ sin 60 ∘ ∣ i j k 6 1 − 1 1 1 0 ∣ = i ( 1 ) − j ( 1 ) + k ( 5 ) = i − j + 5 k ∣ a × b ∣ = 1 + 1 + 25 = 27 ∣ c − a ∣ = 2 2 c 2 + a 2 − 2 a c = 8 c 2 − 12 c + 30 = 0 c = 2 12 + 24 = 6 + 6 ⇒ ∣ ( a × b ) × c ∣ = 27 × ( 6 + 6 ) × 2 3 = 2 a ( 6 + 6 ) Answer the question to unlock