Vector Algebra Questions and Answers

Answer: A

\(\theta\) is \(\frac{\pi}{3}\).

Square the length of \(\vec{a} - \vec{b}\), using \(|\vec{a}| = |\vec{b}| = 1\):

\[ \begin{aligned} |\vec{a} - \vec{b}|^{2} &= 1 + 1 - 2\cos\theta = 1 \\ \cos\theta &= \frac{1}{2} \end{aligned} \]

So \(\theta = \frac{\pi}{3}\): the two unit vectors and their difference form an equilateral triangle. \(\frac{2\pi}{3}\) is the answer when the sum \(\vec{a} + \vec{b}\) is a unit vector, and \(\frac{\pi}{2}\) would make the difference \(\sqrt{2}\) long.

Answer: B

A suitable unit vector is \(\frac{2i - 2j - k}{3}\).

\(\vec{a} + \vec{b} = 3i + 3j\) and \(\vec{a} - \vec{b} = -i + j - 4k\). A vector perpendicular to both is their cross product:

\[ \begin{aligned} (3, 3, 0) \times (-1, 1, -4) &= (-12, 12, 6) \\ &= -6(2, -2, -1) \end{aligned} \]

Its length is \(6 \times 3\), so the unit vectors are \(\pm\frac{2i - 2j - k}{3}\). Check: \((2, -2, -1) \cdot (3, 3, 0) = 0\) and \((2, -2, -1) \cdot (-1, 1, -4) = -2 - 2 + 4 = 0\). \(\frac{2i - 2j + k}{3}\) fails the second check, giving -8.

Answer: D

\(\vec{a} \cdot \vec{b}\) equals 0.

Going round the triangle, \(\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}\), so \(\vec{a} = -(\vec{b} + \vec{c})\). With \(|\vec{b}| = CA = 3\):

\[ \begin{aligned} \vec{a} \cdot \vec{b} &= -(\vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{b}) \\ &= -(9 - 9) = 0 \end{aligned} \]

So BC is perpendicular to CA: the triangle is right-angled at C, which fits BC = AC = 3 and AB = \(3\sqrt{2}\). 9 drops the \(\vec{c} \cdot \vec{b}\) term (keeping only \(|\vec{b}|^{2}\)), and 27 and the \(3\sqrt{3}\) values do not follow from \(\vec{a} = -(\vec{b} + \vec{c})\) at all.

Answer: D

The area is \(\frac{\sqrt{65}}{2}\).

The area of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a} \times \vec{b}|\). With \(\vec{a} = (-1, \frac{1}{4}, 2)\) and \(\vec{b} = (1, \frac{1}{4}, 2)\):

\[ \begin{aligned} \vec{a} \times \vec{b} &= \left(\tfrac{1}{2} - \tfrac{1}{2}\right)\hat{i} - (-2 - 2)\hat{j} \\ &\quad + \left(-\tfrac{1}{4} - \tfrac{1}{4}\right)\hat{k} \\ &= 0\hat{i} + 4\hat{j} - \tfrac{1}{2}\hat{k} \\ |\vec{a} \times \vec{b}| &= \sqrt{16 + \tfrac{1}{4}} \\ &= \frac{\sqrt{65}}{2} \end{aligned} \]

\(\frac{\sqrt{65}}{4}\) is the area of the triangle on those sides, half the parallelogram. \(\sqrt{65}\) forgets that \(16 + \frac{1}{4} = \frac{65}{4}\).

Answer: A

The projection is \(\frac{\sqrt{14}}{2}\).

The scalar projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\):

\[ \begin{aligned} \vec{a} \cdot \vec{b} &= 3(1) + (-1)(2) + (-2)(-3) = 7 \\ |\vec{b}| &= \sqrt{1 + 4 + 9} = \sqrt{14} \\ \frac{7}{\sqrt{14}} &= \frac{7\sqrt{14}}{14} \\ &= \frac{\sqrt{14}}{2} \end{aligned} \]

\(|\vec{a}|\) also happens to be \(\sqrt{14}\), so dividing by the wrong magnitude still gives the same number here; but \(\sqrt{14}\) on its own (option C) forgets the dot product altogether, and \(\frac{14}{\sqrt{2}}\) inverts the fraction.

Answer: B

\(\lambda + \mu = 7\).

The mid-point has position vector \(\frac{\vec{a} + \vec{b}}{2}\). Compare components with \(3\hat{i} + 2\hat{j} + 4\hat{k}\):

\[ \begin{aligned} \tfrac{2 + \mu}{2} &= 3 \Rightarrow \mu = 4 \\ \tfrac{-\lambda + 7}{2} &= 2 \Rightarrow \lambda = 3 \\ \tfrac{5 + 3}{2} &= 4 \text{ (consistent)} \end{aligned} \]

So \(\lambda + \mu = 3 + 4 = 7\). Taking the j-component of \(\vec{a}\) as \(+\lambda\) instead of \(-\lambda\) gives \(\lambda = -3\) and a sum of 1, which is not offered.

Answer: C

\(\cos\theta = \frac{5}{8}\).

For unit vectors, \(|\hat{a}| = |\hat{b}| = 1\), so

\[ \begin{aligned} |\hat{a} - \hat{b}|^{2} &= 1 + 1 - 2\cos\theta \\ \tfrac{3}{4} &= 2 - 2\cos\theta \\ \cos\theta &= \tfrac{5}{8} \end{aligned} \]

Using \(|\hat{a} + \hat{b}|\) instead gives \(\cos\theta = -\frac{5}{8}\), not offered. The option \(\frac{3}{8}\) is \(1 - \cos\theta\).

Answer: D

\(\lambda = -8\).

Collinear vectors are scalar multiples, so their corresponding components are in the same ratio:

\[ \begin{aligned} \frac{4}{-6} &= \frac{\lambda}{12} \\ &= \frac{-6}{9} \\ &= -\frac{2}{3} \end{aligned} \]

\[ \lambda = 12 \times \left(-\tfrac{2}{3}\right) = -8 \]

The i- and k-components agree on the ratio \(-\frac{2}{3}\), which confirms the vectors can be collinear. The option 8 drops the minus sign.

Answer: A

The distance is \(\sqrt{29}\).

The position vector of \(Q\) is that of \(P\) plus \(\overrightarrow{PQ}\):

\[ \begin{aligned} Q &= (2 + 3,\ 1 - 1,\ -3 + 5) = (5, 0, 2) \\ OQ &= \sqrt{25 + 0 + 4} = \sqrt{29} \end{aligned} \]

Subtracting \(\overrightarrow{PQ}\) from \(P\) instead gives \((-1, 2, -8)\) and a distance of \(\sqrt{69}\), one of the distractors.

Answer: A

\(|\vec{a} + \vec{b}|\) = 14.

The angle between the vectors is 85° − 25° = 60°.

\[ \begin{aligned} |\vec{a} + \vec{b}|^{2} &= 36 + 100 + 2(6)(10)\cos 60^{\circ} \\ &= 136 + 60 = 196 \\ |\vec{a} + \vec{b}| &= 14 \end{aligned} \]

Using the angle 110° (the sum of the two angles) or forgetting the cross term would give values not offered. Subtracting instead of adding the cross term gives \(\sqrt{76}\), the magnitude of \(\vec{a} - \vec{b}\).