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NEET-UG and JEE Main Physics · Class 11

Scalar and vector products — NEET & JEE Questions

Scalar and vector products is a topic in Kinematics, a Class 11 Physics chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 15 published NEET-UG and JEE Main questions on Scalar and vector products, each with the marking scheme of the real paper.

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Scalar and vector products MCQs with solutions

10 of the 15 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    The scalar product of two vectors A and B is

    • A. ABcos⁡θAB\cos\theta
    • B. ABsin⁡θAB\sin\theta
    • C. A+BA+B
    • D. AB/cos⁡θAB/\cos\theta
  2. Question 2 · difficulty L2 · understanding

    If two vectors P→=i^+2mj^+mk^\overrightarrow P = \widehat i + 2m\widehat j + m\widehat k and Q→=4i^−2j^+mk^\overrightarrow Q = 4\widehat i - 2\widehat j + m\widehat k are perpendicular to each other. Then, the value of m will be :

    • A. −1-1
    • B. 3
    • C. 1
    • D. 2
  3. Question 3 · difficulty L2 · understanding

    A→\overrightarrow A is a vector quantity such that ∣A→∣|\overrightarrow A | = non-zero constant. Which of the following expression is true for A→\overrightarrow A ?

    • A. A→ . A→=0\overrightarrow A \,.\,\overrightarrow A = 0
    • B. A→×A→<0\overrightarrow A \times \overrightarrow A < 0
    • C. A→×A→=0\overrightarrow A \times \overrightarrow A = 0
    • D. A→×A→>0\overrightarrow A \times \overrightarrow A > 0
  4. Question 4 · difficulty L2 · understanding

    If A→\overrightarrow A and B→\overrightarrow B are two vectors satisfying the relation A→\overrightarrow A . B→\overrightarrow B = ∣A→×B→∣\left| {\overrightarrow A \times \overrightarrow B } \right|. Then the value of ∣A→−B→∣\left| {\overrightarrow A - \overrightarrow B } \right| will be :

    • A. A2+B2+2AB\sqrt {{A^2} + {B^2} + \sqrt 2 AB}
    • B. A2+B2\sqrt {{A^2} + {B^2}}
    • C. A2+B2−2AB\sqrt {{A^2} + {B^2} - \sqrt 2 AB}
    • D. A2+B2+2AB\sqrt {{A^2} + {B^2} + 2AB}
  5. Question 5 · difficulty L2 · understanding

    For three vectors A⃗=(−xi^−6j^−2k^),B⃗=(−i^+4j^+3k^)\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k}), \vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k}) and C⃗=(−8i^−j^+3k^)\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k}), if A⃗⋅(B⃗×C⃗)=0\vec{A} \cdot(\vec{B} \times \vec{C})=0, then value of xx is ________.

  6. Question 6 · difficulty L2 · understanding

    Vectors ai^+bj^+k^a\widehat i + b\widehat j + \widehat k and 2i^−3j^+4k^2\widehat i - 3\widehat j + 4\widehat k are perpendicular to each other when 3a+2b=73a + 2b = 7, the ratio of aa to bb is x2{x \over 2}. The value of xx is ____________.

  7. Question 7 · difficulty L2 · understanding

    If P→×Q→=Q→×P→\overrightarrow P \times \overrightarrow Q = \overrightarrow Q \times \overrightarrow P , the angle between P→\overrightarrow P and Q→\overrightarrow Q is θ\theta(0∘^\circ < θ\theta < 360∘^\circ). The value of 'θ\theta' will be ___________∘^\circ.

  8. Question 8 · difficulty L3 · understanding

    Two particles are located at equal distance from origin. The position vectors of those are represented by A⃗=2i^+3nj^+2k^\vec{A}=2 \hat{i}+3 n \hat{j}+2 \hat{k} and Bˉ=2i^−2j^+4pk^\bar{B}=2 \hat{i}-2 \hat{j}+4 p \hat{k}, respectively. If both the vectors are at right angle to each other, the value of n−1n^{-1} is ________ .

  9. Question 9 · difficulty L3 · understanding

    The resultant of two vectors A⃗\vec{A} and B⃗\vec{B} is perpendicular to A⃗\vec{A} and its magnitude is half that of B⃗\vec{B}. The angle between vectors A⃗\vec{A} and B⃗\vec{B} is _________∘^\circ.

  10. Question 10 · difficulty L3 · understanding

    If a⃗\vec{a} and b⃗\vec{b} makes an angle cos⁡−1(59)\cos ^{-1}\left(\frac{5}{9}\right) with each other, then ∣a⃗+b⃗∣=2∣a⃗−b⃗∣|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}| for ∣a⃗∣=n∣b⃗∣|\vec{a}|=n|\vec{b}| The integer value of n\mathrm{n} is _________.

Common mistakes in Scalar and vector products

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Uses the product of magnitudes without the cosine factor.
  • Uses cosine instead of sine for the cross-product magnitude.
  • Uses $Fs\sin\theta$, which corresponds to a cross-product-type factor.
  • Treats zero dot product as meaning one vector must be zero.

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