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GRE Vector Algebra — practice questions

Explore 23 free Vector Algebra questions in GRE 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 1–23 of 23.

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 →Which of the following is a **vector** quantity?Read the answer and worked solution →The magnitude of the zero (null) vector $\vec 0$ is:Read the answer and worked solution →A unit vector along a non-zero vector $\vec a$ is given by:Read the answer and worked solution →$\overrightarrow{PQ} + \overrightarrow{QR}$ equals (triangle law of addition):Read the answer and worked solution →The magnitude of the vector $\vec a = 4\hat i + 0\hat j + 3\hat k$ is:Read the answer and worked solution →The position vector of the **midpoint** of the segment joining points with position vectors $\vec a$ and $\vec b$ is:Read the answer and worked solution →The triangle inequality for vectors states:Read the answer and worked solution →If $\vec{PQ} = \vec a$, then $\vec{QP}$ equals:Read the answer and worked solution →The position vector of the **centroid** of triangle with vertices $A(\vec a),\ B(\vec b),\ C(\vec c)$ is:Read the answer and worked solution →Two non-zero vectors $\vec a$ and $\vec b$ are **collinear** (parallel) if and only if there exists a non-zero scalar $m$ such that:Read the answer and worked solution →Vector $\vec a$ is directed due north with $|\vec a| = 24$ and vector $\vec b$ is directed due west with $|\vec b| = 7$. The magnitude $|\vec a + \vec b|$ equals:Read the answer and worked solution →The distance from the point $P(2, 3, 4)$ to the **x-axis** is:Read the answer and worked solution →The point dividing the line segment joining $A(2, -6, 8)$ and $B(-1, 3, -4)$ internally in ratio $1:3$ has coordinates:Read the answer and worked solution →If $\vec a = 4\hat i + 3\hat k$ and $\vec b = -2\hat i + \hat j + 5\hat k$, then $2\vec a + 5\vec b$ equals:Read the answer and worked solution →Vectors $\vec a = \hat i - 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 6\hat j + 9\hat k$ are:Read the answer and worked solution →The position vector of the centroid of the **tetrahedron** with vertices $A(\vec a), B(\vec b), C(\vec c), D(\vec d)$ is:Read the answer and worked solution →If vectors $2\hat i - q\hat j + 3\hat k$ and $4\hat i - 5\hat j + 6\hat k$ are collinear, then $q$ equals:Read the answer and worked solution →Vector $\vec{AB} = 2\hat i - 4\hat j + 7\hat k$ has initial point $A(1, 5, 0)$. The terminal point $B$ has coordinates:Read the answer and worked solution →Find a unit vector in the direction of $\vec a = \hat i - 2\hat j$: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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