Vector Algebra — JEE Main practice

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Q1 · 2026

The velocity of a particle is given as v⃗=−𝑥i^+2𝑦j^−𝑧k^\vec{v} = -𝑥 \hat{i} + 2𝑦 \hat{j} - 𝑧 \hat{k} m/s. The magnitude of acceleration at point (1,2,4)(1, 2, 4) is ________ m/s2^2.

  • A.

    6\sqrt{6}

  • B.

    33\sqrt{33}

  • C.

    00

  • D.

    99

Answer: D

  1. The acceleration of a particle is the time derivative of its velocity:

    a⃗=dv⃗dt\vec{a}=\frac{d\vec{v}}{dt}

  2. The velocity components are:

    vx=−𝑥,vy=2𝑦,vz=−𝑧v_x=-𝑥,\qquad v_y=2𝑦,\qquad v_z=-𝑧

  3. Differentiate each component with respect to time using the chain rule:

    ax=dvxdt=dvxd𝑥⋅d𝑥dt=(−1)(−𝑥)=𝑥a_x=\frac{dv_x}{dt}=\frac{dv_x}{d𝑥}\cdot\frac{d𝑥}{dt}=(-1)(-𝑥)=𝑥 ay=dvydt=dvyd𝑦⋅d𝑦dt=2(2𝑦)=4𝑦a_y=\frac{dv_y}{dt}=\frac{dv_y}{d𝑦}\cdot\frac{d𝑦}{dt}=2(2𝑦)=4𝑦 az=dvzdt=dvzd𝑧⋅d𝑧dt=(−1)(−𝑧)=𝑧a_z=\frac{dv_z}{dt}=\frac{dv_z}{d𝑧}\cdot\frac{d𝑧}{dt}=(-1)(-𝑧)=𝑧

  4. Hence, the acceleration vector is:

    a⃗=𝑥i^+4𝑦j^+𝑧k^\vec{a}=𝑥\hat{i}+4𝑦\hat{j}+𝑧\hat{k}

  5. At the point (1,2,4)(1,2,4):

    a⃗=i^+8j^+4k^\vec{a}=\hat{i}+8\hat{j}+4\hat{k}

  6. Its magnitude is:

∣a⃗∣=12+82+42=1+64+16=81=9 m/s2.\begin{aligned} |\vec{a}| &= \sqrt{1^2+8^2+4^2} \\ &= \sqrt{1+64+16} \\ &= \sqrt{81} \\ &= 9\ \text{m/s}^2. \end{aligned}
  1. Therefore, the magnitude of acceleration is 9 m/s29\ \text{m/s}^2, so the correct option is D.

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