Vector Algebra Questions and Answers

Answer: A

The correct match is (P) → (5), (Q) → (4), (R) → (2), (S) → (1).

\(MM^{T} = I\) makes \(M\) orthogonal, so its rows are orthonormal and so are its columns, and \(\vec{u}, \vec{v}, \vec{w}\) are exactly the three columns.

Row 1 has unit length: \(\alpha^{2} + \frac{1}{2} + \frac{1}{2} = 1\), so \(\alpha = 0\). Rows 1 and 2 are perpendicular: \(\frac{\beta}{\sqrt{2}} - \frac{1}{\sqrt{6}} = 0\), so \(\beta = \frac{1}{\sqrt{3}}\). Then the unit-length columns give:

\[ \begin{aligned} \gamma^{2} &= 1 - 0 - \tfrac{1}{3} = \tfrac{2}{3} \\ \delta^{2} &= 1 - \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6} \\ \gamma^{2} + \delta^{2} &= \tfrac{5}{6} \end{aligned} \]

(Q) In an orthonormal basis the coefficient of \(\vec{u}\) is the dot product with \(\vec{u}\): \(x = \vec{u} \cdot \hat{j} = \frac{1}{\sqrt{3}}\). (R) \(\vec{u} \cdot (\vec{v} \times \vec{w}) = \det M = \pm 1\), so its modulus is 1. (S) \(\vec{v} \times \vec{w}\) is a unit vector perpendicular to both, which is \(\pm\vec{u}\), so \(\vec{u} \times (\vec{v} \times \vec{w}) = \vec{0}\) and its modulus is 0.

The fourth option swaps R and S, the third gives \(x\) the value \(\frac{1}{\sqrt{2}}\), and the second misplaces P as well.

Answer: A

The value is 614.

Add \(\vec{a}\) and \(\vec{b}\) component by component, then take the dot product with \(\vec{c} = 11\hat{i} + 0\hat{j} - 17\hat{k}\):

\[ \begin{aligned} \vec{a} + \vec{b} &= -6\hat{i} - 4\hat{j} - 40\hat{k} \\ (\vec{a} + \vec{b}) \cdot \vec{c} &= (-6)(11) + (-40)(-17) \\ &= -66 + 680 = 614 \end{aligned} \]

The \(\hat{j}\) term contributes nothing because \(\vec{c}\) has no \(\hat{j}\) component. 746 comes from adding 66 instead of subtracting it; 2 and 134 do not follow from the components.

Answer: B

An acceptable value is x = 2.

Rotation does not change length, so doubling gives \(|\vec{B}| = 2|\vec{A}|\), that is \(|\vec{B}|^{2} = 4|\vec{A}|^{2}\).

\[ \begin{aligned} 16 + (4x - 2)^{2} + 4 &= 4(1 + x^{2} + 9) \\ 16x^{2} - 16x + 24 &= 4x^{2} + 40 \\ 3x^{2} - 4x - 4 &= 0 \end{aligned} \]

\[ \begin{aligned} (x - 2)(3x + 2) &= 0 \\ x &= 2 \text{ or } -\tfrac{2}{3} \end{aligned} \]

Check x = 2: \(|\vec{A}|^{2} = 14\) and \(|\vec{B}|^{2} = 16 + 36 + 4 = 56 = 4 \times 14\). The values 1, 3 and 4/3 do not satisfy the equation.

Answer: C

The area of triangle \(OPR\) is \(\frac{3\sqrt{3}}{2}\).

Perpendicularity gives \(\vec{a}\cdot(\vec{a} - \vec{b}) = 0\), so \(|\vec{a}|^{2} = \vec{a}\cdot\vec{b}\). Expand the two given lengths:

\[ \begin{aligned} |\vec{a}|^{2} + |\vec{b}|^{2} + 2\,\vec{a}\cdot\vec{b} &= 21 \\ |\vec{a}|^{2} + |\vec{b}|^{2} - 2\,\vec{a}\cdot\vec{b} &= 9 \end{aligned} \]

Subtracting, \(4\,\vec{a}\cdot\vec{b} = 12\), so \(\vec{a}\cdot\vec{b} = 3\) and \(|\vec{a}|^{2} = 3\). Adding, \(|\vec{a}|^{2} + |\vec{b}|^{2} = 15\), so \(|\vec{b}|^{2} = 12\).

With \(P\) at \(\vec{a}\) and \(R\) at \(\vec{a} + \vec{b}\), the area is half the cross product, and \(\vec{a} \times \vec{a} = \vec{0}\):

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2}|\vec{a} \times (\vec{a} + \vec{b})| \\ &= \tfrac{1}{2}|\vec{a} \times \vec{b}| \end{aligned} \]

\[ \begin{aligned} |\vec{a} \times \vec{b}|^{2} &= |\vec{a}|^{2}|\vec{b}|^{2} \\ &\quad - (\vec{a}\cdot\vec{b})^{2} \\ &= 36 - 9 = 27 \\ \text{Area} &= \tfrac{1}{2}\sqrt{27} = \tfrac{3\sqrt{3}}{2} \end{aligned} \]

\(\sqrt{3}\) is \(|\vec{a}|\) itself, and \(\frac{3}{2}\) is half the dot product rather than half the cross product; \(\frac{\sqrt{3}}{2}\) is half of \(|\vec{a}|\).

Answer: E

\(|\vec{a}| = \) \(\sqrt{194}\).

Perpendicular vectors have zero dot product:

\[ \begin{aligned} (\vec{a} + \vec{b})\cdot(\vec{a} - \vec{b}) &= 2\lambda + 128 - 18\lambda = 0 \\ \lambda &= 8 \end{aligned} \]

Adding the two given vectors and halving gives \(\vec{a}\):

\[ \begin{aligned} \vec{a} &= \tfrac{1}{2}(10\hat{i} + 24\hat{j} - 10\hat{k}) = 5\hat{i} + 12\hat{j} - 5\hat{k} \\ |\vec{a}| &= \sqrt{25 + 144 + 25} = \sqrt{194} \end{aligned} \]

The other surds come from a wrong \(\lambda\) or from taking \(|\vec{a} + \vec{b}|\) instead of \(|\vec{a}|\).

Answer: B

The angle is \(\frac{\pi}{6}\).

The projection of \(\vec{a}\) on \(\vec{b}\) is \(|\vec{a}|\cos\theta\):

\[ \begin{aligned} 12\cos\theta &= 6\sqrt{3} \\ \cos\theta &= \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{6} \end{aligned} \]

\(\frac{\pi}{3}\) has cosine \(\frac{1}{2}\), which would need a projection of 6. A positive projection rules out the obtuse angles \(\frac{2\pi}{3}\) and \(\frac{3\pi}{4}\), and a right angle would give zero projection.

Answer: E

\(|\vec{b}| = \) \(\frac{7}{2}\).

Being parallel, \(\vec{b} = k\vec{a}\). Since \(|\vec{a}| = \sqrt{36 + 4 + 9} = 7\):

\[ \begin{aligned} \vec{a}\cdot\vec{b} &= k|\vec{a}|^{2} = 49k = \frac{49}{2} \\ k &= \frac{1}{2},\quad |\vec{b}| = \frac{1}{2}(7) = \frac{7}{2} \end{aligned} \]

7 is \(|\vec{a}|\) itself, 49 is \(|\vec{a}|^{2}\), and 14 doubles instead of halving.

Answer: A

The value is 3.

For any vectors, \(|\vec{a} + \vec{b}|^{2} - |\vec{a} - \vec{b}|^{2} = 4\,\vec{a}\cdot\vec{b}\). Find \(\vec{a}\cdot\vec{b}\) from the given length, using \(|\vec{a}| = |\vec{b}| = 1\):

\[ \begin{aligned} |\vec{a} + \vec{b}|^{2} &= 2 + 2\,\vec{a}\cdot\vec{b} = \frac{14}{4} \\ \vec{a}\cdot\vec{b} &= \frac{3}{4} \\ 4\,\vec{a}\cdot\vec{b} &= 3 \end{aligned} \]

Check: \(|\vec{a} - \vec{b}|^{2} = 2 - \frac{3}{2} = \frac{1}{2}\), and \(\frac{7}{2} - \frac{1}{2} = 3\). 7 and \(\sqrt{7}\) come from mishandling \(\left(\frac{\sqrt{14}}{2}\right)^{2}\).

Answer: D

\(BD =\) \(2\sqrt{3}\).

\(\vec{BD} = \vec{AD} - \vec{AB}\):

\[ \begin{aligned} \vec{BD} &= (3 - 5)\hat{i} + (2 - 4)\hat{j} + (-1 + 3)\hat{k} \\ &= -2\hat{i} - 2\hat{j} + 2\hat{k} \\ |\vec{BD}| &= \sqrt{4 + 4 + 4} = 2\sqrt{3} \end{aligned} \]

The other diagonal, \(|\vec{AB} + \vec{AD}| = |8\hat{i} + 6\hat{j} - 4\hat{k}| = 2\sqrt{29}\), is not offered; \(3\sqrt{3}\), \(4\sqrt{3}\), \(2\sqrt{5}\) and \(3\sqrt{2}\) do not match.

Answer: B

\(\alpha =\) 4.

The projection of \(\vec{a}\) on \(\vec{c}\) is \(\frac{\vec{a}\cdot\vec{c}}{|\vec{c}|}\), with \(|\vec{c}| = \sqrt{1 + 4 + 4} = 3\):

\[ \begin{aligned} \vec{a}\cdot\vec{c} &= \alpha + 6 - 4 = \alpha + 2 \\ \frac{\alpha + 2}{3} &= 2 \;\Rightarrow\; \alpha = 4 \end{aligned} \]

Forgetting to divide by \(|\vec{c}|\) gives \(\alpha = 0\); the other values do not make the projection 2.