Vector Algebra Questions and Answers

Answer: E

The angle is \(\frac{\pi}{3}\).

Expand \(|\vec{a} - \vec{b}|^{2}\):

\[ \begin{aligned} 49 &= 64 + 25 - 2(8)(5)\cos\theta \\ 80\cos\theta &= 40 \;\Rightarrow\; \cos\theta = \tfrac{1}{2} \end{aligned} \]

so \(\theta = \frac{\pi}{3}\). \(\frac{2\pi}{3}\) is what \(|\vec{a} + \vec{b}| = 7\) would give, the sign trap; \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\) and \(\frac{\pi}{6}\) do not satisfy \(\cos\theta = \frac{1}{2}\).

Answer: E

The value is 0.

The product is a difference of squares, and \(\vec{a}\cdot\vec{a} = |\vec{a}|^{2}\):

\[ \begin{aligned} (2\vec{a} + 3\vec{b})\cdot(2\vec{a} - 3\vec{b}) &= 4|\vec{a}|^{2} - 9|\vec{b}|^{2} \\ &= 4(9) - 9(4) = 0 \end{aligned} \]

The cross terms \(\pm 6\,\vec{a}\cdot\vec{b}\) cancel, so the angle between the vectors is not needed. Options \(\pm 6\) and \(\pm 12\) come from using \(|\vec{a}|\) instead of \(|\vec{a}|^{2}\) or mixing the coefficients.

Answer: D

The statement that is not true is \(m(\vec{a}.\vec{b}) = m\vec{a}.m\vec{b}\).

A scalar can be taken out of a dot product only once: \((m\vec{a}).(m\vec{b}) = m^{2}(\vec{a}.\vec{b})\). So the right side is m times too large unless m is 0 or 1.

The other statements are standard laws: vector addition is associative; scalar multiplication distributes over a sum of vectors and over a sum of scalars; and \(m(\vec{a} \times \vec{b}) = (m\vec{a}) \times \vec{b}\), because the scalar is applied to one factor of the cross product, not both.

Answer: C

The value is 108.

Each scalar comes out of the dot product.

\[ \begin{aligned} (3\vec{a}).(3\vec{b}) &= 9(\vec{a}.\vec{b}) \\ &= 9 \times 12 = 108 \end{aligned} \]

36 takes the factor 3 out only once, and 144 squares the dot product instead of the scalar. 4 and 16 divide instead of multiplying.

Answer: C

The value is 8.

The dot product distributes like ordinary multiplication, and the cross terms cancel because \(\vec{a}.\vec{b} = \vec{b}.\vec{a}\).

\[ \begin{aligned} (\vec{a} + \vec{b}).(\vec{a} - \vec{b}) &= |\vec{a}|^{2} - |\vec{b}|^{2} \\ &= (9 + 4 + 4) - (1 + 4 + 4) \\ &= 17 - 9 = 8 \end{aligned} \]

Working directly gives the same: \(\vec{a} + \vec{b} = 4\hat{i} + 4\hat{j}\) and \(\vec{a} - \vec{b} = 2\hat{i} + 4\hat{k}\), whose dot product is 8 + 0 + 0 = 8. The neighbouring values come from an arithmetic slip in one of the squared lengths.

Answer: A

The conclusion is \(\vec{a} - \vec{d}\) is parallel to \(\vec{b} - \vec{c}\).

Two non-zero vectors are parallel when their cross product is zero. Expand \((\vec{a} - \vec{d}) \times (\vec{b} - \vec{c})\).

\[ \begin{aligned} &\vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c} \\ &= \vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d} \end{aligned} \]

The given conditions make the first and last terms cancel (\(\vec{a} \times \vec{b} = \vec{c} \times \vec{d}\)) and the middle two cancel (\(\vec{a} \times \vec{c} = \vec{b} \times \vec{d}\)), so the cross product is zero. The other pairings do not reduce to these two conditions.

Answer: D

\(\sin\theta = \frac{\sqrt{17}}{9}\).

Find the cosine from the dot product:

\[ \begin{aligned} \vec{AB}\cdot\vec{AC} &= -2 + 20 + 22 = 40 \\ |\vec{AB}| &= \sqrt{4 + 100 + 121} = 15, \ |\vec{AC}| = 3 \\ \cos\theta &= \frac{40}{45} = \frac{8}{9} \end{aligned} \]

Then \(\sin\theta = \sqrt{1 - \frac{64}{81}} = \frac{\sqrt{17}}{9}\). The other square roots come from arithmetic slips in the magnitudes or the dot product, and \(\frac{4}{9}\) is half the cosine.

Answer: C

\(\vec{a} + \vec{b} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})\).

Let \(\vec{c} = 2\hat{i} + 3\hat{j} + 4\hat{k}\). Since \(\vec{c} \times \vec{b} = -\vec{b} \times \vec{c}\), the condition becomes

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

so \(\vec{a} + \vec{b}\) is parallel to \(\vec{c}\). Its magnitude \(\sqrt{29}\) equals \(|\vec{c}| = \sqrt{4 + 9 + 16}\), so \(\vec{a} + \vec{b} = \pm\vec{c}\).

The vectors with a changed sign on one component are not parallel to \(\vec{c}\), and \(\pm\sqrt{29}\,\vec{c}\) has magnitude 29.

Answer: E

The scalar triple product is 0.

\(\vec{b} = 2\hat{i} + 4\hat{j} + 8\hat{k} = 2\vec{a}\), so \(\vec{a}\) and \(\vec{b}\) are parallel and \(\vec{a} \times \vec{b} = \vec{0}\). The dot product of the zero vector with \(\vec{c}\) is 0, whatever \(\vec{c}\) is.

Equivalently, the determinant formed by the three vectors has two proportional rows. The non-zero options come from working out the cross product with an arithmetic slip.

General rule: the scalar triple product is zero whenever two of the three vectors are parallel, or more generally whenever the three are coplanar.

Answer: E

The bisecting unit vector is \(\frac{i + j}{\sqrt{2}}\).

The internal bisector of two vectors lies along the sum of their unit vectors. Both given vectors have length 3, since \(\sqrt{4 + 1 + 4} = \sqrt{1 + 4 + 4} = 3\):

\[ \begin{aligned} \frac{2i + j + 2k}{3} + \frac{i + 2j - 2k}{3} &= \frac{3i + 3j}{3} \\ &= i + j \end{aligned} \]

Normalising, \(\frac{i + j}{\sqrt{2}}\). The difference of the unit vectors, \(\frac{i - j + 4k}{3}\), gives the external bisector, and none of the \(\sqrt{3}\) options is parallel to i + j.