Exams Joint Entrance Examination Mathematics Vector Algebra Question 100 Question 100 of 379 Medium NUMERICAL
Let a ⃗ = i ^ + j ^ + k ^ , b → = 2 i ^ + 2 j ^ + k ^ \vec{a}=\hat{i}+\hat{j}+\hat{k},\overset{\to }{b}=2\hat{i}+2\hat{j}+\hat{k} a = i ^ + j ^ + k ^ , b → = 2 i ^ + 2 j ^ + k ^ and d → = a ⃗ × b → \overset{\to }{d}=\vec{a}\times \overset{\to }{b} d → = a × b → . If c → \overset{\to }{c} c → is a vector such that a ⃗ ⋅ c → = ∣ c → ∣ \vec{a}\cdot \overset{\to }{c}=|\overset{\to }{c}| a ⋅ c → = ∣ c → ∣ , ∣ c → − 2 a ⃗ ∣ 2 = 8 |\overset{\to }{c}-2\vec{a}{|}^{2}=8 ∣ c → − 2 a ∣ 2 = 8 and the angle between d → \overset{\to }{d} d → and c → \overset{\to }{c} c → is π 4 \frac{\pi }{4} 4 π , then ∣ 10 − 3 b → ← ⋅ c → ∣ + ∣ d → × c → ∣ 2 |10-3\overset{\to \leftarrow }{b}\cdot \overset{\to }{c}|+|\overset{\to }{d}\times \overset{\to }{c}{|}^{2} ∣10 − 3 b →← ⋅ c → ∣ + ∣ d → × c → ∣ 2 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 6
Solutions · 1 Post your solution Official solution
a → = i ^ + j ^ + k ^ b → ← = 2 i ^ + 2 j ^ + k ^ d → ← = a → × b → = − i ^ + j ^ ∣ c → − 2 a → ∣ 2 = 8 ∣ c ∣ 2 + 4 ∣ a ∣ 2 − 4 ( a ⋅ c ) = 8 c 2 + 12 − 4 c = 8 c 2 − 4 c + 4 = 0 ∣ c ∣ = 2 d → ← = a → × b → d → ← × c → = ( a → × b → ) 2 × c → ( ∣ d ∣ ∣ c ∣ sin π 4 ) 2 = ( ( a ⋅ c ) ⋅ b − ( b ⋅ c ) ⋅ a ) 2 4 = 4 b 2 + ( b ⋅ c ) 2 2 ( a ) 2 − 2 ( b ⋅ c ) ( a ⋅ b ) \begin{gathered} \overset{\to }{a}=\hat{i}+\hat{j}+\hat{k} \\ \overset{\to \leftarrow }{b}=2\hat{i}+2\hat{j}+\hat{k} \\ \overset{\to \leftarrow }{d}=\overset{\to }{a}\times \overset{\to }{b} \\ =-\hat{i}+\hat{j} \\ |\overset{\to }{c}-2\overset{\to }{a}{|}^{2}=8 \\ |c{|}^{2}+4|a{|}^{2}-4(a\cdot c)=8 \\ {c}^{2}+12-4c=8 \\ {c}^{2}-4c+4=0 \\ |c|=2 \\ \overset{\to \leftarrow }{d}=\overset{\to }{a}\times \overset{\to }{b} \\ \overset{\to \leftarrow }{d}\times \overset{\to }{c}=(\overset{\to }{a}\times \overset{\to }{b}{)}^{2}\times \overset{\to }{c} \\ {(|d||c|\sin \frac{\pi }{4})}^{2}=((a\cdot c)\cdot b-(b\cdot c)\cdot a{)}^{2} \\ 4=4{b}^{2}+(b\cdot c{)}^{2}2(a{)}^{2}-2(b\cdot c)(a\cdot b) \end{gathered} a → = i ^ + j ^ + k ^ b →← = 2 i ^ + 2 j ^ + k ^ d →← = a → × b → = − i ^ + j ^ ∣ c → − 2 a → ∣ 2 = 8 ∣ c ∣ 2 + 4∣ a ∣ 2 − 4 ( a ⋅ c ) = 8 c 2 + 12 − 4 c = 8 c 2 − 4 c + 4 = 0 ∣ c ∣ = 2 d →← = a → × b → d →← × c → = ( a → × b → ) 2 × c → ( ∣ d ∣∣ c ∣ sin 4 π ) 2 = (( a ⋅ c ) ⋅ b − ( b ⋅ c ) ⋅ a ) 2 4 = 4 b 2 + ( b ⋅ c ) 2 2 ( a ) 2 − 2 ( b ⋅ c ) ( a ⋅ b )
b . c = x 4 = 36 + 3 x 2 − 20 x 3 x 2 − 20 x + 32 = 0 3 x 2 − 12 x − 8 x + 32 = 0 x = 8 3 , 4 = 8 3 , 4 = 8 3 ∣ 10 − 3 b . c ∣ + ∣ d × c ∣ 2 ∣ 10 − 8 ∣ + ( 2 ) 2 ⇒ 6 \begin{gathered} b.c=x \\ 4=36+3{x}^{2}-20x \\ 3{x}^{2}-20x+32=0 \\ 3{x}^{2}-12x-8x+32=0 \\ x=\frac{8}{3},4 \\ =\frac{8}{3},4 \\ =\frac{8}{3} \\ |10-3b.c|+|d\times c{|}^{2} \\ |10-8|+(2{)}^{2} \\ \Rightarrow 6 \end{gathered} b . c = x 4 = 36 + 3 x 2 − 20 x 3 x 2 − 20 x + 32 = 0 3 x 2 − 12 x − 8 x + 32 = 0 x = 3 8 , 4 = 3 8 , 4 = 3 8 ∣10 − 3 b . c ∣ + ∣ d × c ∣ 2 ∣10 − 8∣ + ( 2 ) 2 ⇒ 6 Answer the question to unlock