Mathematics · Vector Algebra

JEE Main 2024 — 6 April, Shift 2 — Question 17

Let a⃗=2i^+j^−k^,b⃗=((a⃗×(i^+j^))×i^)×i^\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}.

Then the square of the projection of a⃗\vec{a} on b⃗\vec{b} is :

  1. Option A:

    15\frac{1}{5}

  2. Option B:

    22

    Correct
  3. Option C:

    13\frac{1}{3}

  4. Option D:

    ) 23\frac{2}{3}

Answer: B

Step-by-step solution

a⃗×(i^ +j^ )=∣i^j^ k^ 21−1110∣\vec{a}\times\left(\overset{{}}{\mathop{{\hat{i}}}}\,+\overset{{}}{\mathop{{\hat{j}}}}\, \right)=\left| \begin{matrix}{\hat{i}} & \overset{{}}{\mathop{{\hat{j}}}}\, & \overset{{}}{\mathop{{\hat{k}}}}\, \\2 & 1 & -1 \\1 & 1 & 0 \\\end{matrix} \right|

=i^−j^+k^=\hat{i}-\hat{j}+\hat{k}

(a⃗×(i^×j^))×i^=k^+j^(\vec{a} \times(\hat{i} \times \hat{j})) \times \hat{i}=\hat{k}+\hat{j} ((a⃗×(i^×j^))×i)×i^=j^−k^((\vec{a} \times(\hat{i} \times \hat{j})) \times i) \times \hat{i}=\hat{j}-\hat{k}

projection of a⃗\vec{a}

on b^=a⃗⋅b⃗∣b⃗∣\hat{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

=1+12=2=\frac{1+1}{\sqrt{2}}=\sqrt{2}

the square of the projection of a⃗\vec{a} on b⃗\vec{b} is 22

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Vector Algebra
Topic
Projection & component of a vector along another vector.