JEE Main 2024 8 April Evening Shift
Exam JEE Main
Year 2024
Date 8 April
Shift Evening Shift
Subject Mathematics
Chapter Vector Algebra JEE Main 2024 · 8 April · Evening Shift · Mathematics · Vector Algebra
Let a =4 i - j + k , b =11 i - j + k and c be a vector such that (...
Let
a ⃗ = 4 i ^ − j ^ + k ^ , b ⃗ = 11 i ^ − j ^ + k ^ \vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k} a = 4 i ^ − j ^ + k ^ , b = 11 i ^ − j ^ + k ^ and
c ⃗ \vec{c} c be a vector such that
( a ⃗ + b ⃗ ) × c ⃗ = c ⃗ × ( − 2 a ⃗ + 3 b ⃗ ) (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) ( a + b ) × c = c × ( − 2 a + 3 b ) . If
( 2 a ⃗ + 3 b ⃗ ) ⋅ c ⃗ = 1670 (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 ( 2 a + 3 b ) ⋅ c = 1670 , then
∣ c ⃗ ∣ 2 |\vec{c}|^2 ∣ c ∣ 2 is equal to :
Options A 1609 B 1618 C 1600 D 1627
Step-by-step solution
( a → + b → ) × c → − c → × ( − 2 a → + 3 b → ) = 0 ( a → + b → ) × c → + ( − 2 a → + 3 b → ) × c → = 0 ⇒ ( a → + b → ) − 2 a → + 3 b → ) × c → = 0 ⇒ c → = λ ( 4 b → − a → ) ⇒ = λ ( 44 i ^ − 4 j ^ + 4 k ^ − 4 i ^ + j ^ − k ^ ) = λ ( 40 i ^ − 3 j ^ + 3 k ^ ) \begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}+(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \\ & \Rightarrow=\lambda(44 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =\lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\end{aligned} ( a + b ) × c − c × ( − 2 a + 3 b ) = 0 ( a + b ) × c + ( − 2 a + 3 b ) × c = 0 ⇒ ( a + b ) − 2 a + 3 b ) × c = 0 ⇒ c = λ ( 4 b − a ) ⇒= λ ( 44 i ^ − 4 j ^ + 4 k ^ − 4 i ^ + j ^ − k ^ ) = λ ( 40 i ^ − 3 j ^ + 3 k ^ )
Now
( 8 i ^ − 2 j ^ + 2 k ^ + 33 i ^ − 3 j ^ + 3 k ^ ) ⋅ λ ( 40 i ^ − 3 j ^ + 3 k ^ ) = 1670 ⇒ ( 41 i ^ − 5 j ^ + 5 k ^ ) ⋅ ( 40 i ^ − 3 j ^ + 3 k ^ ) × λ = 1670 ) ⇒ ( 1640 + 15 + 15 ) λ = 1670 ⇒ λ = 1 so c → = 40 i ^ − 3 j ^ − 3 k ^ ⇒ ∣ c → ∣ 2 = 1600 + 9 + 9 = 1618 \begin{aligned}
& (8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+33 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot \lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=1670 \\
& \Rightarrow(41 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times \lambda=1670) \\
& \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 \\
& \text{ so } \overrightarrow{\mathrm{c}}=40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\
& \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618
\end{aligned} ( 8 i ^ − 2 j ^ + 2 k ^ + 33 i ^ − 3 j ^ + 3 k ^ ) ⋅ λ ( 40 i ^ − 3 j ^ + 3 k ^ ) = 1670 ⇒ ( 41 i ^ − 5 j ^ + 5 k ^ ) ⋅ ( 40 i ^ − 3 j ^ + 3 k ^ ) × λ = 1670 ) ⇒ ( 1640 + 15 + 15 ) λ = 1670 ⇒ λ = 1 so c = 40 i ^ − 3 j ^ − 3 k ^ ⇒ ∣ c ∣ 2 = 1600 + 9 + 9 = 1618
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