If the components of a =α i +β j +γ k along and perpendicular to b =3 i + j - k respectively, are (16)/(11)(3 i + j - k ) and (1)/(11)(-4 i -5 j -17 k ) , then α^2+β^2+γ^2 is equal to :

Correct answer: A. 26 let a _ 11 = component of a along b a _1= component of a perpendicular to b aligned & a_ 11 =(16)/(11)(3 i + j - k ) & a_1=(1)/(11)(-4 i -5 j -17 k ) & a = a_ 11 + a_1 aligned aligned & array l a = =(16)/(11)(3 i + j - k )+(1)/(11)(-4 i -5 j -17 k ) 11 j -(33)/(11) k array & aligned α=4 i + j -3 k aligned & α^2+β^2+γ^2=16+1+9=26 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Vector Algebra

If the components of a =α i +β j +γ k along and perpendicular to b...

If the components of

\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}

along and perpendicular to

\overrightarrow{\mathrm{b}}=3 \hat{i}+\hat{j}-\hat{k}

respectively, are

\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})

and

\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})

, then

\alpha^2+\beta^2+\gamma^2

is equal to :

Options

A $26$
B $18$
C $23$
D $16$

Correct answer

A. $26$

Step-by-step solution

let

\vec{a}_{11}=\text{ component

} \vec{a} \text{ along } \vec{b} \vec{a}_1=

component of

\vec{a}

perpendicular to

\vec{b}

\begin{aligned} & \overrightarrow{\mathrm{a}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}_1=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ & \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}_{11}+\overrightarrow{\mathrm{a}_1 \end{aligned}

\begin{aligned} & \begin{array}{l}\therefore \overrightarrow{\mathrm{a}}= \\ \quad=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ \overrightarrow{11} \hat{\mathrm{j}}-\frac{33}{11} \hat{\mathrm{k}}\end{array} \\ & \begin{aligned} \alpha=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}\end{aligned}

$ \\ & \alpha^2+\beta^2+\gamma^2=16+1+9=26\end{aligned}

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