A-Level Maths

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Vectors in the Plane practice questions — A-Level Maths

10 free multiple-choice problems on vectors in the plane, ordered to match A-Level Maths difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Problems & worked solutions

Problem #0425 International

Problem 1Vectors in the Plane

Are the vectors u=i+2j\vec{u} = \vec{i} + 2\vec{j} and v=2i+4j\vec{v} = 2\vec{i} + 4\vec{j} collinear?

Show answer & worked solution
  1. A. Yes, v=2u\vec{v} = 2\vec{u}✓ correct
  2. B. No, the dot product is non-zero
  3. C. Yes, but only because they are perpendicular
  4. D. No, they have different magnitudes

v=2u\vec{v} = 2\vec{u}, so the two vectors are collinear (parallel).

Problem #0426 International

Problem 2Vectors in the Plane

For u=i+j\vec{u} = \vec{i} + \vec{j} and v=ai2j\vec{v} = a\vec{i} - 2\vec{j}, find aRa \in \mathbb{R} so that u\vec{u} and v\vec{v} are collinear.

Show answer & worked solution
  1. A. 2-2✓ correct
  2. B. 12-\dfrac{1}{2}
  3. C. 12\dfrac{1}{2}
  4. D. 22

v=ku\vec{v} = k \vec{u} for some kka=ka = k and 2=k-2 = k. So a=2a = -2.

Problem #0428 International

Problem 3Vectors in the Plane

The vector AB\overrightarrow{AB} from A(1,2)A(1, 2) to B(4,6)B(4, 6) equals:

Show answer & worked solution
  1. A. 5i+8j5\vec{i} + 8\vec{j}
  2. B. 3i+4j3\vec{i} + 4\vec{j}✓ correct
  3. C. 3i4j-3\vec{i} - 4\vec{j}
  4. D. i+2j\vec{i} + 2\vec{j}

AB=(41)i+(62)j=3i+4j\overrightarrow{AB} = (4 - 1)\vec{i} + (6 - 2)\vec{j} = 3\vec{i} + 4\vec{j}.

Problem #0427 International

Problem 4Vectors in the Plane

Find mRm \in \mathbb{R} so that u=mi+3j\vec{u} = m\vec{i} + 3\vec{j} and v=4i+(m+1)j\vec{v} = 4\vec{i} + (m + 1)\vec{j} are perpendicular.

Show answer & worked solution
  1. A. 1-1
  2. B. 37-\dfrac{3}{7}✓ correct
  3. C. 37\dfrac{3}{7}
  4. D. 11

uv=4m+3(m+1)=7m+3=0m=37\vec{u} \cdot \vec{v} = 4m + 3(m + 1) = 7m + 3 = 0 \Rightarrow m = -\dfrac{3}{7}.

Problem #0422 International

Problem 5Vectors in the Plane

For u=i+2j\vec{u} = \vec{i} + 2\vec{j} and v=3ij\vec{v} = 3\vec{i} - \vec{j}, the sum u+v\vec{u} + \vec{v} equals:

Show answer & worked solution
  1. A. 4i+2j4\vec{i} + 2\vec{j}
  2. B. 4i3j4\vec{i} - 3\vec{j}
  3. C. 4i+j4\vec{i} + \vec{j}✓ correct
  4. D. 2i+3j-2\vec{i} + 3\vec{j}

Add components: (1+3)i+(21)j=4i+j(1 + 3)\vec{i} + (2 - 1)\vec{j} = 4\vec{i} + \vec{j}.

Problem #0424 International

Problem 6Vectors in the Plane

The dot product uv\vec{u} \cdot \vec{v} for u=2i+3j\vec{u} = 2\vec{i} + 3\vec{j}, v=i4j\vec{v} = \vec{i} - 4\vec{j} equals:

Show answer & worked solution
  1. A. 10-10✓ correct
  2. B. 6-6
  3. C. 55
  4. D. 1414

uv=21+3(4)=212=10\vec{u} \cdot \vec{v} = 2 \cdot 1 + 3 \cdot (-4) = 2 - 12 = -10.

Problem #0423 International

Problem 7Vectors in the Plane

For u=3i2j\vec{u} = 3\vec{i} - 2\vec{j}, the vector 2u2\vec{u} equals:

Show answer & worked solution
  1. A. 5i2j5\vec{i} - 2\vec{j}
  2. B. 3i4j3\vec{i} - 4\vec{j}
  3. C. 6i4j6\vec{i} - 4\vec{j}✓ correct
  4. D. 6i+4j6\vec{i} + 4\vec{j}

Multiply each component by 22: 2u=6i4j2\vec{u} = 6\vec{i} - 4\vec{j}.

Problem #0421 International

Problem 8Vectors in the Plane

For u=3i+4j\vec{u} = 3\vec{i} + 4\vec{j}, the magnitude u|\vec{u}| equals:

Show answer & worked solution
  1. A. 33
  2. B. 44
  3. C. 55✓ correct
  4. D. 77

u=32+42=25=5|\vec{u}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.

Problem #0430 International

Problem 9Vectors in the Plane

For u=i+j\vec{u} = \vec{i} + \vec{j} and v=ai2j\vec{v} = a\vec{i} - 2\vec{j}, find aRa \in \mathbb{R} so that u+v2=u2+v2|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2.

Show answer & worked solution
  1. A. 2-2
  2. B. 1-1
  3. C. 22✓ correct
  4. D. 44

uv=a+(2)=a2=0a=2\vec{u} \cdot \vec{v} = a + (-2) = a - 2 = 0 \Rightarrow a = 2.

Problem #0429 International

Problem 10Vectors in the Plane

Let a\vec{a} and b\vec{b} be two non-collinear vectors. Find mRm \in \mathbb{R} so that u=3a(m+1)b\vec{u} = 3\vec{a} - (m + 1)\vec{b} and v=(m1)a5b\vec{v} = (m - 1)\vec{a} - 5\vec{b} are collinear.

Show answer & worked solution
  1. A. {4}\{-4\}
  2. B. {4}\{4\}
  3. C. {4,4}\{-4, 4\}✓ correct
  4. D. {2,8}\{-2, 8\}

3(5)=(m1)((m+1))3 \cdot (-5) = (m - 1)\,(-(m+1)), i.e. 15=(m1)(m+1)=1m2-15 = -(m-1)(m+1) = 1 - m^2, so m2=16m^2 = 16 and m{4,4}m \in \{-4, 4\}.

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