Vector Practice Questions

The magnitude of a vector \(\langle a, b \rangle\) is found using \(|\mathbf{v}| = \sqrt{a^2 + b^2}\). Substituting:

\(|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Choice B adds the components instead of squaring them first. Choice C squares and adds correctly but forgets to take the square root.

To add two vectors, add their corresponding components:

\(\mathbf{u} + \mathbf{v} = \langle 2 + (-3),\ -1 + 5 \rangle = \langle -1, 4 \rangle\)

Choice B subtracts the vectors rather than adding them. Choice C gets the \(x\)-component right but subtracts the \(y\)-components instead of adding.

Scalar multiplication distributes to each component:

\(3\mathbf{v} = \langle 3(-2),\ 3(5) \rangle = \langle -6, 15 \rangle\)

Choice B adds 3 to each component instead of multiplying. Choice C multiplies correctly but drops the negative sign on the \(x\)-component.

A unit vector is found by dividing each component by the vector’s magnitude.

First, find \(|\mathbf{v}|\):

\(|\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)

Then divide each component by 10 and simplify:

\(\hat{\mathbf{v}} = \left\langle \dfrac{6}{10},\ \dfrac{8}{10} \right\rangle = \left\langle \dfrac{3}{5},\ \dfrac{4}{5} \right\rangle\)

Choice C takes the reciprocal of each component rather than dividing by the magnitude. Choice D swaps the two components.

The dot product multiplies corresponding components and adds the results:

\(\mathbf{u} \cdot \mathbf{v} = (2)(4) + (3)(-1) = 8 + (-3) = 5\)

Choice B adds all four components together without multiplying pairs. Choice C gets the sign wrong by treating −3 as positive.

Two vectors are parallel if one is a scalar multiple of the other. Check each choice against \(\mathbf{v} = \langle 2, -4 \rangle\):

Two vectors are orthogonal if and only if their dot product equals 0.

Compute \(\mathbf{u} \cdot \mathbf{v}\):

\(\mathbf{u} \cdot \mathbf{v} = (3)(-4) + (4)(3) = -12 + 12 = 0\)

Since the dot product is 0, the vectors are orthogonal. Choice C confuses the condition — equal magnitudes have nothing to do with orthogonality. Choice D describes a condition for non-parallel vectors, which is a completely separate concept.

The angle \(\theta\) between two vectors satisfies \(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\). Compute each part:

\(\mathbf{u} \cdot \mathbf{v} = (1)(1) + (1)(0) = 1\)

\(|\mathbf{u}| = \sqrt{1^2 + 1^2} = \sqrt{2},\)\(\: |\mathbf{v}| = \sqrt{1^2 + 0^2} = 1\)

\(\cos \theta = \dfrac{1}{\sqrt{2} \cdot 1} = \dfrac{1}{\sqrt{2}} \: \Rightarrow \: \theta = 45°\)

Choice D would result from a dot product of 0, which would indicate orthogonal vectors — that is not the case here.

A vector with magnitude \(r\) and direction angle \(\theta\) has component form \(\langle r\cos\theta,\ r\sin\theta \rangle\). Substituting \(r = 10\) and \(\theta = 30°\):

\(\left\langle 10\cos 30°,\ 10\sin 30° \right\rangle\)\(\:= \left\langle 10 \cdot \dfrac{\sqrt{3}}{2},\ 10 \cdot \dfrac{1}{2} \right\rangle = \langle 5\sqrt{3},\ 5 \rangle\)

Choice B swaps the sine and cosine values. Choice C uses \(\sin 45° = \cos 45°\) as if the angle were 45°. Choice D corresponds to a direction angle of 0°.

Apply the formula with \(\mathbf{u} = \langle 4, 3 \rangle\) and \(\mathbf{v} = \langle 1, 0 \rangle\):

\(\mathbf{u} \cdot \mathbf{v} = (4)(1) + (3)(0) = 4\)

\(|\mathbf{v}|^2 = 1^2 + 0^2 = 1\)

\(\text{proj}_{\mathbf{v}}\,\mathbf{u} = \dfrac{4}{1} \cdot \langle 1, 0 \rangle = \langle 4, 0 \rangle\)

Since \(\mathbf{v}\) points along the \(x\)-axis, projecting \(\mathbf{u}\) onto it keeps only the \(x\)-component. Choice B is the component of \(\mathbf{u}\) perpendicular to \(\mathbf{v}\), not the projection onto it. Choice C returns the original vector unchanged.

Subtract the corresponding components of \(\mathbf{v}\) from \(\mathbf{u}\):

\(\mathbf{u} \:- \mathbf{v} = \langle 5 – 3,\ -2 – 4 \rangle = \langle 2, -6 \rangle\)

Choice B adds the vectors instead of subtracting. Choice C reverses the order and computes \(\mathbf{v} − \mathbf{u}\). Choice D gets the \(x\)-component right but treats \(-2-4\) as positive 6.

For two vectors to be parallel, their components must be proportional. Set up the ratio and solve:

\(\dfrac{k}{2} = \dfrac{9}{6} = \dfrac{3}{2}\)

\(k = 3\)

Choice D comes from multiplying \(9 \times 2\) rather than finding a proportional value. Choice C is the denominator of the simplified ratio, not the solution for \(k\).

The two velocity vectors point in perpendicular directions (east and north), so their resultant magnitude is found using the Pythagorean theorem:

\(|\mathbf{r}| = \sqrt{30^2 + 40^2} = \sqrt{900 + 1{,}600}\)\(\: = \sqrt{2{,}500} = 50\)

Choice B simply adds the two speeds, which would only be valid if both velocities were in the same direction. Choice C subtracts them. This is a classic 3-4-5 Pythagorean triple scaled by 10.

The direction angle is measured counterclockwise from the positive \(x\)-axis. Since the \(x\)-component is negative and the \(y\)-component is positive, the vector lies in Quadrant II.

Find the reference angle first:

\(\alpha = \arctan\left(\dfrac{|3|}{|-3|}\right) = \arctan(1) = 45°\)

For a Quadrant II vector, subtract the reference angle from 180°:

\(\theta = 180° – 45° = 135°\)

Choice B is only the reference angle and ignores the quadrant. Choice C would place the vector in Quadrant III, and Choice D in Quadrant IV.

Set \(a\mathbf{u} + b\mathbf{v} = \mathbf{w}\) and write a system using the components:

\(a(1) + b(2) = 5 \: \Rightarrow \: a + 2b = 5\)

\(a(2) + b(-1) = 0 \: \Rightarrow \: 2a – b = 0\)

From the second equation, \(b = 2a\). Substituting into the first:

\(a + 2(2a) = 5 \: \Rightarrow \: 5a = 5 \: \Rightarrow \: a = 1,\ b = 2\)

So \(\mathbf{w} = \mathbf{u} + 2\mathbf{v}\).