Let a =2 i - j - k , b = i +3 j - k and c =2 i + j +3 k . Let v be the vector in the plane of the vectors a and b , such that the length of its projection on the vector c is 1 14 . Then | v | is equal to

Correct answer: B. 35 2 v = λ a + μ b = (2λ+μ,\ -λ+3μ,\ -λ-μ) . v · c = 2(2λ+μ) + (-λ+3μ) + 3(-λ-μ) = 2μ . | c | = 4+1+9 = 14 . Projection length = |2μ| 14 = 1 14 |μ| = (1)/(2) . | v |^2 = 6λ^2 + 11μ^2 = 6(1) + (11)/(4) = (35)/(4) . | v | = 35 2 .

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Vector Algebra

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

Let

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

and

\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}

. Let

\vec{v}

be the vector in the plane of the vectors

\vec{a}

and

\vec{b}

, such that the length of its projection on the vector

\vec{c}

\frac{1}{\sqrt{14}}

. Then

|\vec{v}|

is equal to

Options

A13
B $\frac{\sqrt{35}}{2}$
C $\frac{\sqrt{21}}{2}$
D7

Correct answer

B. $\frac{\sqrt{35}}{2}$

Step-by-step solution

\vec{v} = \lambda\vec{a} + \mu\vec{b} = (2\lambda+\mu,\ -\lambda+3\mu,\ -\lambda-\mu)

\vec{v} \cdot \vec{c} = 2(2\lambda+\mu) + (-\lambda+3\mu) + 3(-\lambda-\mu) = 2\mu

|\vec{c}| = \sqrt{4+1+9} = \sqrt{14}

.
Projection length

= \frac{|2\mu|}{\sqrt{14}} = \frac{1}{\sqrt{14}} \Rightarrow |\mu| = \frac{1}{2}

|\vec{v}|^2 = 6\lambda^2 + 11\mu^2 = 6(1) + \frac{11}{4} = \frac{35}{4}

|\vec{v}| = \frac{\sqrt{35}}{2}

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