Vector Subtraction and Multiplication by a Scalar Quiz

Vector Subtraction and Multiplication by a Scalar

This quiz tests Lesson 5 of Motion in a Plane. You will practise subtracting vectors by adding the negative vector, drawing the triangle and parallelogram views, working with components, and scaling vectors by positive and negative scalars.

What to expect

20 medium-difficulty questions on subtraction, negative vectors, and scalar multiplication.
Questions on magnitude, direction, unit vectors, and properties of scalar multiplication.
Applications such as relative velocity and reversing a particle’s course.

Remember the two key ideas: ##\vec{A} – \vec{B} = \vec{A} + (-\vec{B})##, and multiplying by a scalar changes length while keeping or flipping the line of action.

Q1. Vector subtraction ##\vec{A} - \vec{B}## is best defined as:

##\vec{A} + (-\vec{B})##

##\vec{B} - \vec{A}##

##\vec{A} + \vec{B}##

##-(\vec{A} + \vec{B})##

Subtraction is adding the negative vector: flip ##\vec{B}## to ##-\vec{B}## and add it head to tail to ##\vec{A}##.

Q2. The negative of a nonzero vector ##\vec{B}## has:

The same magnitude and the same direction

The same magnitude and the opposite direction

A different magnitude and the same direction

Zero magnitude

The negative of a nonzero vector has the same magnitude and the opposite direction.

Q3. Which equation expresses the additive inverse property of a vector?

##\vec{B} - \vec{B} = \vec{B}##

##\vec{B} + (-\vec{B}) = \vec{0}##

##\vec{B} + \vec{B} = \vec{0}##

##-\vec{B} = \vec{B}##

A vector plus its negative gives the zero vector, which has no magnitude and no defined direction.

Q4. A displacement of 5 m east is reversed. Its negative is:

5 m east

5 m north

5 m west

0 m

Reversing direction through 180 degrees turns 5 m east into 5 m west, with the same magnitude.

Q5. When ##\vec{A}## and ##\vec{B}## share a common tail, the vector ##\vec{A} - \vec{B}## points:

From the head of ##\vec{A}## to the head of ##\vec{B}##

From the head of ##\vec{B}## to the head of ##\vec{A}##

From the common tail to the head of ##\vec{A}##

From the common tail to the head of ##\vec{B}##

In the triangle view with a common tail, the difference runs from the head of the subtracted vector to the head of the first vector.

Q6. To compute ##\vec{A} - \vec{B}## graphically, you should:

Draw ##-\vec{B}## and add it head to tail to ##\vec{A}##

Draw ##\vec{B}## and add it head to tail to ##\vec{A}##

Reverse ##\vec{A}## and add it to ##\vec{B}##

Draw both vectors from a common tail and join their heads

Graphical subtraction is reversal followed by addition: flip ##\vec{B}##, then use the head-to-tail rule.

Q7. If ##\vec{A} = 3\hat{i} + 4\hat{j}## and ##\vec{B} = \hat{i} + 2\hat{j}##, then ##\vec{A} - \vec{B}## equals:

##2\hat{i} + 2\hat{j}##

##4\hat{i} + 6\hat{j}##

##2\hat{i} + 6\hat{j}##

##-2\hat{i} - 2\hat{j}##

Subtract components: ##(3-1)\hat{i} + (4-2)\hat{j} = 2\hat{i} + 2\hat{j}##.

Q8. For ##\vec{A} = 5\hat{i} - 3\hat{j}## and ##\vec{B} = 2\hat{i} + \hat{j}##, the vector ##\vec{B} - \vec{A}## is:

##3\hat{i} - 4\hat{j}##

##-3\hat{i} + 4\hat{j}##

##7\hat{i} - 2\hat{j}##

##3\hat{i} + 4\hat{j}##

##\vec{B} - \vec{A} = (2-5)\hat{i} + (1-(-3))\hat{j} = -3\hat{i} + 4\hat{j}##.

Q9. Multiplying a nonzero vector by a positive scalar ##k > 1##:

Reverses its direction and keeps its magnitude

Keeps its direction and increases its magnitude

Keeps its direction and decreases its magnitude

Makes its magnitude zero

For a nonzero vector, multiplying by a positive scalar greater than one preserves its direction and multiplies its magnitude by ##k##, increasing it.

Q10. Multiplying a nonzero vector by a negative scalar:

Keeps the direction unchanged

Reverses the direction and scales the magnitude

Always gives the zero vector

Changes the line of action to a perpendicular line

A negative scalar reverses a nonzero vector's direction and multiplies its magnitude by the absolute value of the scalar.

Q11. If ##\vec{A}## has magnitude 6 units, the magnitude of ##-3\vec{A}## is:

6 units

3 units

18 units

2 units

The magnitude scales by the absolute value of the scalar: ##\left|-3\right| \times 6 = 18## units.

Q12. A unit vector in the direction of a nonzero vector ##\vec{A}## is obtained by:

Multiplying ##\vec{A}## by its magnitude

Dividing ##\vec{A}## by its magnitude

Adding ##\vec{A}## to itself

Taking the negative of ##\vec{A}##

Dividing a vector by its magnitude gives a vector of length one in the same direction.

Q13. Which statement about the zero vector is correct?

It has magnitude zero and a well-defined direction

It has magnitude one and no direction

It has magnitude zero and no defined direction

It has magnitude zero and points east

The zero vector has magnitude zero and no defined direction; it is the additive identity.

Q14. Which of the following is a valid property of scalar multiplication?

##k(\vec{A} + \vec{B}) = k\vec{A} + k\vec{B}##

##k(\vec{A} + \vec{B}) = k\vec{A} + \vec{B}##

##(k + m)\vec{A} = k\vec{A} + m\vec{B}##

##k\vec{A} = \vec{A}## for every scalar ##k##

Scalar multiplication distributes over vector addition, so the scalar multiplies each term.

Q15. If ##\vec{A} = 4\hat{i} - 2\hat{j}##, then ##2\vec{A}## equals:

##8\hat{i} - 4\hat{j}##

##2\hat{i} - \hat{j}##

##8\hat{i} + 4\hat{j}##

##-8\hat{i} + 4\hat{j}##

Each component is doubled: ##2(4)\hat{i} + 2(-2)\hat{j} = 8\hat{i} - 4\hat{j}##.

Q16. A particle moving with velocity ##\vec{v}## reverses its course with the same speed. Its new velocity is:

##2\vec{v}##

##\vec{v}##

##-\vec{v}##

##\vec{0}##

Reversing course with the same speed means multiplying the velocity by ##-1##, giving ##-\vec{v}##.

Q17. The vector ##\vec{A} - \vec{A}## is equal to:

##2\vec{A}##

##\vec{A}##

##-\vec{A}##

##\vec{0}##

A vector minus itself is the zero vector, since ##\vec{A} + (-\vec{A}) = \vec{0}##.

Q18. Relative velocity of body 1 with respect to body 2 is computed as:

##\vec{v}_1 + \vec{v}_2##

##\vec{v}_1 - \vec{v}_2##

##\vec{v}_2 - \vec{v}_1##

##\vec{v}_1 \times \vec{v}_2##

Relative velocity is the difference of the two velocity vectors, obtained by adding the negative of the second.

Q19. If ##\vec{A} = 2\hat{i} + 3\hat{j}## and ##\vec{B} = 2\hat{i} + 3\hat{j}##, then ##\vec{A} - \vec{B}## is:

##4\hat{i} + 6\hat{j}##

##\vec{0}##

##-4\hat{i} - 6\hat{j}##

##\hat{i} + \hat{j}##

Equal vectors have equal components, so their difference is the zero vector.

Q20. Multiplying a vector by the scalar 0 gives:

The same vector

A unit vector

The zero vector

The negative of the vector

Any vector multiplied by zero has zero magnitude, producing the zero vector.