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MP DTE Vector Algebra — practice questions

Explore 60 free Vector Algebra questions in MP DTE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Vector Algebra questions and exercises with answers

Showing questions 1–30 of 60.

The dot product (scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is:Read the answer and worked solution →Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if:Read the answer and worked solution →The cross product $\vec{a} \times \vec{b}$ of two vectors has magnitude:Read the answer and worked solution →Two vectors of magnitudes $3$ and $4$ are perpendicular to each other. The magnitude of their resultant is:Read the answer and worked solution →A quantity having only magnitude (no direction) is called:Read the answer and worked solution →Vector addition is:Read the answer and worked solution →The dot product ā · b̄ is:Read the answer and worked solution →The cross product ā × b̄ is:Read the answer and worked solution →What is î × ĵ?Read the answer and worked solution →Two vectors are PERPENDICULAR iff:Read the answer and worked solution →Two vectors are PARALLEL (or anti-parallel) iff:Read the answer and worked solution →The MAGNITUDE of a × b equals:Read the answer and worked solution →Find magnitude of ā = 3î + 4ĵ:Read the answer and worked solution →Find ā · b̄ if ā = 2î + 3ĵ and b̄ = î - 2ĵ:Read the answer and worked solution →Find ā × b̄ where ā = î + 2ĵ + k̂ and b̄ = 2î + ĵ + 3k̂:Read the answer and worked solution →Angle between two vectors ā = 2î + 2ĵ and b̄ = 4î - 4ĵ:Read the answer and worked solution →Find unit vector in direction of ā = 3î + 4ĵ:Read the answer and worked solution →Area of parallelogram formed by ā and b̄ equals:Read the answer and worked solution →Position vectors of P and Q are p̄ and q̄. The point R divides PQ internally in ratio m:n. Position vector of R:Read the answer and worked solution →Three vectors ā, b̄, c̄ are coplanar iff their SCALAR TRIPLE PRODUCT [ā b̄ c̄] equals:Read the answer and worked solution →For vectors ā = î + ĵ and b̄ = î - ĵ, find |ā + b̄|:Read the answer and worked solution →Projection of ā = 3î + 4ĵ on b̄ = î + ĵ:Read the answer and worked solution →If ā × b̄ = 0 and neither is the zero vector, then:Read the answer and worked solution →Volume of parallelepiped formed by ā, b̄, c̄:Read the answer and worked solution →For unit vectors î, ĵ, k̂: î · ĵ + ĵ · k̂ + k̂ · î =Read the answer and worked solution →For vectors ā = î - ĵ + 2k̂ and b̄ = 3î + 2ĵ - k̂, find the AREA of the parallelogram formed (use |ā × b̄|):Read the answer and worked solution →Find cos θ for ā = 2î + 3ĵ - 4k̂ and b̄ = î + 2ĵ + 3k̂ (where θ is the angle between them):Read the answer and worked solution →Find scalar triple product [î ĵ k̂]:Read the answer and worked solution →Vectors ā = î + 2ĵ + 3k̂ and b̄ = 2î + 3ĵ + αk̂ are perpendicular. Find α:Read the answer and worked solution →For vectors ā = î - ĵ + 2k̂ and b̄ = 3î + 2ĵ - k̂, find the area of triangle formed:Read the answer and worked solution →

Worked practice questions

The dot product (scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is:

  1. $|\vec{a}||\vec{b}|\cos\theta$
  2. $|\vec{a}|^2 + |\vec{b}|^2$
  3. $|\vec{a}||\vec{b}|\sin\theta$
  4. $|\vec{a}||\vec{b}|\tan\theta$
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Correct answer: A. $|\vec{a}||\vec{b}|\cos\theta$

The dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ where $\theta$ is the angle between them. The result is a scalar. The sine version (option A) defines the magnitude of the cross product, which is a vector, not a scalar. Mixing up the two is a recurring exam trap.

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Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if:

  1. $\vec{a} \times \vec{b} = 0$
  2. $\vec{a} \cdot \vec{b} = 0$
  3. $\vec{a} + \vec{b} = 0$
  4. $|\vec{a}| = |\vec{b}|$
Show answer and solution

Correct answer: B. $\vec{a} \cdot \vec{b} = 0$

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$. For two non-zero vectors, this is zero exactly when $\cos\theta = 0$, i.e. $\theta = 90°$. Option B describes *parallel* vectors (cross product zero). Option D would require the vectors to be exact opposites of each other, not just perpendicular.

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The cross product $\vec{a} \times \vec{b}$ of two vectors has magnitude:

  1. $|\vec{a}||\vec{b}|\cos\theta$
  2. $|\vec{a}||\vec{b}|\sin\theta$
  3. $|\vec{a}| + |\vec{b}|$
  4. $|\vec{a}|^2 - |\vec{b}|^2$
Show answer and solution

Correct answer: B. $|\vec{a}||\vec{b}|\sin\theta$

$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$. The cross product is a vector perpendicular to both $\vec{a}$ and $\vec{b}$, with direction given by the right-hand rule. Its magnitude equals the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$. Geometric reading: when $\theta = 0$ or $\pi$ (parallel or anti-parallel), $\sin\theta = 0$ and the cross product vanishes, which is why parallel vectors have zero cross product.

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