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JEE Main Vector Algebra — practice questions

Explore 34 free Vector Algebra questions in JEE Main 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 34.

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 →Multiplying a vector by a scalar gives a result that is:Read the answer and worked solution →A vector multiplied by a scalar stays collinear with the:Read the answer and worked solution →A negative scalar multiplier gives a vector whose direction is:Read the answer and worked solution →The magnitude of a scaled vector equals the original magnitude times the scalar's:Read the answer and worked solution →Writing a vector as x i plus y j plus z k gives its:Read the answer and worked solution →The numbers x, y and z in component form are also called:Read the answer and worked solution →The section formula concerns a segment PQ divided by a third point in how many ways?Read the answer and worked solution →A point dividing PQ between P and Q divides it:Read the answer and worked solution →A point dividing PQ beyond the segment divides it:Read the answer and worked solution →The projection of a vector on a line is itself a:Read the answer and worked solution →The magnitude of the projection uses the cosine of the angle with the:Read the answer and worked solution →The direction of the projection depends on the sign of the:Read the answer and worked solution →Projection of a vector on a line is the dot product with the line's:Read the answer and worked solution →The scalar product obeys laws that are:Read the answer and worked solution →The scalar product is an operation that is:Read the answer and worked solution →The vector product is an operation that is:Read the answer and worked solution →The angle between two vectors can be found in terms of their:Read the answer and worked solution →The cross product of two nonzero vectors is zero when they are:Read the answer and worked solution →For mutually perpendicular unit vectors, i cross j equals:Read the answer and worked solution →The area of a parallelogram with sides a and b uses the magnitude of their:Read the answer and worked solution →The area of a triangle from its three vertices can be found using:Read the answer and worked solution →The result a cross b is a:Read the answer and worked solution →True or false on vector algebraRead the answer and worked solution →Quick picks on vector algebraRead the answer and worked solution →Match each idea to its meaningRead the answer and worked solution →Dot product or cross productRead 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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