Exams Joint Entrance Examination Mathematics Vector Algebra Question 107 Question 107 of 379 Medium NUMERICAL
Let a ⃗ = 3 i ^ + 2 j ^ + k ^ , b ⃗ = 2 i ^ − j ^ + 3 k ^ \vec{a}=3\hat{i}+2\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k} a = 3 i ^ + 2 j ^ + k ^ , b = 2 i ^ − j ^ + 3 k ^ and c ⃗ \vec{c} c be a vector such that ( a ⃗ + b ⃗ ) × c ⃗ = 2 ( a ⃗ × b ⃗ ) + 24 j ^ − 6 k ^ (\vec{a}+\vec{b})\times \vec{c}=2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k} ( a + b ) × c = 2 ( a × b ) + 24 j ^ − 6 k ^ and ( a ⃗ − b ⃗ + i ^ ) ⋅ c ⃗ = − 3 (\vec{a}-\vec{b}+\hat{i})\cdot \vec{c}=-3 ( a − b + i ^ ) ⋅ c = − 3 . Then ∣ c ⃗ ∣ 2 |\vec{c}{|}^{2} ∣ c ∣ 2 is equal to \_\_\_\_\_\_\_\_.
Answer: 38
Solutions · 1 Post your solution Official solution
( a → + b → ) × c → = 2 ( a → × b → ) + 24 j ^ − 6 k ^ ( 5 i ^ + j ^ + 4 k ^ ) × c → = 2 ( 7 i ^ − 7 j ^ − 7 k ^ ) + 24 j ^ − 6 k ^ ∣ i ^ j ^ k ^ 5 1 4 x y z ∣ = 14 i ^ + 10 j ^ − 20 k ^ ⇒ i ^ ( z − 4 y ) − j ^ ( 5 z − 4 x ) + k ^ ( 5 y − x ) = 14 i ^ + 10 j ^ − 20 k ^ z − 4 y = 14 , 4 x − 5 z = 10 , 5 y − x = − 20 ( a − b + i ) ⋅ c → = − 3 ( 2 i ^ + 3 j ^ − 2 k ^ ) ⋅ c → = − 3 2 x + 3 y − 2 z = − 3 ∴ x = 5 , y = − 3 , z = 2 ∣ c → ∣ 2 = 25 + 9 + 4 = 38 \begin{gathered} (\overset{\to }{a}+\overset{\to }{b})\times \overset{\to }{c}=2(\overset{\to }{a}\times \overset{\to }{b})+24\hat{j}-6\hat{k} \\ (5\hat{i}+\hat{j}+4\hat{k})\times \overset{\to }{c}=2(7\hat{i}-7\hat{j}-7\hat{k})+24\hat{j}-6\hat{k} \\ |\begin{array}{|c|}\hat{i} \quad \hat{j} \quad \hat{k} \\ 5 \quad 1 \quad 4 \\ x \quad y \quad z\end{array}|=14\hat{i}+10\hat{j}-20\hat{k} \\ \Rightarrow \hat{i}(z-4y)-\hat{j}(5z-4x)+\hat{k}(5y-x)=14\hat{i}+10\hat{j}-20\hat{k} \\ z-4y=14,4x-5z=10,5y-x=-20 \\ (a-b+i)\cdot \overset{\to }{c}=-3 \\ (2\hat{i}+3\hat{j}-2\hat{k})\cdot \overset{\to }{c}=-3 \\ 2x+3y-2z=-3 \\ \therefore x=5,y=-3,z=2 \\ |\overset{\to }{c}{|}^{2}=25+9+4=38 \end{gathered} ( a → + b → ) × c → = 2 ( a → × b → ) + 24 j ^ − 6 k ^ ( 5 i ^ + j ^ + 4 k ^ ) × c → = 2 ( 7 i ^ − 7 j ^ − 7 k ^ ) + 24 j ^ − 6 k ^ ∣ i ^ j ^ k ^ 5 1 4 x y z ∣ = 14 i ^ + 10 j ^ − 20 k ^ ⇒ i ^ ( z − 4 y ) − j ^ ( 5 z − 4 x ) + k ^ ( 5 y − x ) = 14 i ^ + 10 j ^ − 20 k ^ z − 4 y = 14 , 4 x − 5 z = 10 , 5 y − x = − 20 ( a − b + i ) ⋅ c → = − 3 ( 2 i ^ + 3 j ^ − 2 k ^ ) ⋅ c → = − 3 2 x + 3 y − 2 z = − 3 ∴ x = 5 , y = − 3 , z = 2 ∣ c → ∣ 2 = 25 + 9 + 4 = 38 Answer the question to unlock