Let a = i +2 j +3 k , b =3 i + j - k and c be three vectors such that c is coplanar with a and b . If the vector C is perpendicular to b and a · c =5 , then | c | is equal to

Correct answer: A. (11)/(6) aligned & c =λ( b ×( a × b )) & =λ(( b · b ) a -( a · b ) b ) & =λ(11 a -2 b )=λ(11 i+22 j+33 k-6 i-2 j+2 k) & =λ(5 i+20 j+35 k) & =5 λ(5 i+4 j+7 k) & = Given c · a =5 & =5 λ(1+8+21)=5 =λ=(1)/(30) & c =(1)/(6)(i+4 j+7 k) & c = 1+16+49 6 = (11)/(6) aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Vector Algebra

Let a = i +2 j +3 k , b =3 i + j - k and c be three vectors such...

Let

\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+\hat{j}-\hat{k}

and

\vec{c}

be three vectors such that

\vec{c}

is coplanar with

\vec{a}

and

\vec{b}

. If the vector

\vec{C}

is perpendicular to

\vec{b}

and

\vec{a} \cdot \vec{c}=5

, then

|\vec{c}|

is equal to

Options

A $\sqrt{\frac{11}{6}}$
B $\frac{1}{3 \sqrt{2}}$
C $16$
D $18$

Correct answer

A. $\sqrt{\frac{11}{6}}$

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{b}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})) \\ & =\lambda((\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{b}}) \\ & =\lambda(11 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{ b}})=\lambda(11 \mathrm{i}+22 \mathrm{j}+33 \mathrm{k}-6 \mathrm{i}-2 \mathrm{j}+2 \mathrm{k}) \\ & =\lambda(5 \mathrm{i}+20 \mathrm{j}+35 \mathrm{k}) \\ & =5 \lambda(5 \mathrm{i}+4 \mathrm{j}+7 \mathrm{k}) \\ & =\text{ Given } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=5 \\ & =5 \lambda(1+8+21)=5 \quad=\lambda=\frac{1}{30} \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\frac{1}{6}(\mathrm{i}+4 \mathrm{j}+7 \mathrm{k}) \\ & \overrightarrow{\mathrm{c}}=\frac{\sqrt{1+16+49}}{6}=\sqrt{\frac{11}{6}}\end{aligned}

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