Vector Algebra Questions and Answers

Answer: E

The angle between \(\vec{a}\) and \(\vec{c}\) is \(\dfrac{\pi}{6}\).

From \(\vec{a} - \vec{b} + \vec{c} = 0\), \(\vec{b} = \vec{a} + \vec{c}\). Square both sides:

\[ \begin{aligned} |\vec{b}|^{2} &= |\vec{a}|^{2} + |\vec{c}|^{2} + 2|\vec{a}||\vec{c}|\cos\theta \\ 13 + 6\sqrt{3} &= 4 + 9 + 12\cos\theta \\ \cos\theta &= \frac{\sqrt{3}}{2} \end{aligned} \]

So \(\theta = \frac{\pi}{6}\). \(\frac{\pi}{3}\) would need \(\cos\theta = \frac{1}{2}\), that is \(|\vec{b}|^{2} = 19\), and \(\frac{\pi}{2}\) would need \(|\vec{b}|^{2} = 13\).

Answer: D

\(\vec{a}\) equals \(\pm\dfrac{1}{3\sqrt{3}}(\vec{b} \times \vec{c})\).

\(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), so it is parallel to \(\vec{b} \times \vec{c}\): \(\vec{a} = \lambda(\vec{b} \times \vec{c})\). Compare lengths:

\[ \begin{aligned} |\vec{b} \times \vec{c}| &= 3 \times 4 \times \sin\tfrac{\pi}{3} = 6\sqrt{3} \\ |\lambda| \times 6\sqrt{3} &= 2 \Rightarrow |\lambda| = \frac{1}{3\sqrt{3}} \end{aligned} \]

The sign is not fixed, since either direction along the perpendicular works. Using \(\cos\frac{\pi}{3}\) instead of the sine, or dropping a factor, gives the other coefficients.

Answer: C

The value is 25.

Square the condition \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) by taking its dot product with itself:

\[ \begin{aligned} |\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2S &= 0 \\ 9 + 16 + 25 + 2S &= 0 \\ S &= -25 \end{aligned} \]

where \(S = \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}\). Its absolute value is 25.

50 is the sum of squares before halving, which is not offered; 20, 24, 30 and 35 do not come from this identity.

Answer: E

The values are \(k = \pm 7\).

For any two vectors, \(|\vec{a} \times \vec{b}|^{2} + (\vec{a}.\vec{b})^{2}\) equals \(|\vec{a}|^{2}|\vec{b}|^{2}\), because the two terms are \(|\vec{a}|^{2}|\vec{b}|^{2}\sin^{2}\theta\) and \(|\vec{a}|^{2}|\vec{b}|^{2}\cos^{2}\theta\) (Lagrange's identity).

\[ \begin{aligned} k^{2} &= |\vec{a}|^{2}|\vec{b}|^{2} = 49 \times 1 \\ k &= \pm 7 \end{aligned} \]

The identity holds whatever the angle \(\theta\), so \(k^{2} = 49\) exactly and both signs are allowed. 5, 3 and their negatives do not square to 49.

Answer: D

\(\vec{OC} = \hat{i} - \hat{j}\).

The midpoint of \(A(-2, 1)\) and \(B(4, -3)\) is

\[ \begin{aligned} C &= \left(\frac{-2 + 4}{2}, \frac{1 + (-3)}{2}\right) \\ &= (1, -1) \end{aligned} \]

so its position vector from the origin is \(\hat{i} - \hat{j}\). \(-\hat{i} + \hat{j}\) has both signs reversed, \(\hat{i} + \hat{j}\) gets the y-coordinate's sign wrong, and \(\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}\) halves the wrong sums.

Answer: B

The projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{1}{\sqrt{3}}\).

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

\[ \begin{aligned} \vec{a}.\vec{b} &= 1 - 1 + 1 = 1 \\ |\vec{b}| &= \sqrt{1 + 1 + 1} = \sqrt{3} \\ \text{projection} &= \frac{1}{\sqrt{3}} \end{aligned} \]

\(\frac{1}{3}\) divides by \(|\vec{b}|^{2}\) instead of \(|\vec{b}|\), and \(\sqrt{3}\) is \(|\vec{b}|\) itself. The dot product is positive, so the negative options are wrong.

Answer: C

x lies in \((-\infty, 1) \cup (2, \infty)\).

The angle between two vectors is acute when their dot product is positive.

\[ \begin{aligned} \vec{a}\cdot\vec{b} &= x^{2} - 3x + 2 \\ &= (x - 1)(x - 2) \gt 0 \end{aligned} \]

The product is positive when both factors have the same sign: \(x \lt 1\) or \(x \gt 2\). The other options use the wrong roots, such as -1 or 0, which come from sign slips in \(-3x\).

Answer: D

The possible values are \(\pm 6\).

Let \(\vec{v} = \hat{i} - \hat{j} + \hat{k}\). Since \(\vec{v} \times \vec{b} = -\vec{b} \times \vec{v}\), the condition becomes \((\vec{a} + \vec{b}) \times \vec{v} = \vec{0}\), so \(\vec{a} + \vec{b}\) is parallel to \(\vec{v}\).

\[ \begin{aligned} \vec{a} + \vec{b} &= \lambda(\hat{i} - \hat{j} + \hat{k}) \\ |\lambda|\sqrt{3} &= 3\sqrt{3} \Rightarrow \lambda = \pm 3 \\ (\vec{a} + \vec{b})\cdot(3, 2, 1) &= \lambda(3 - 2 + 1) = \pm 6 \end{aligned} \]

Forgetting that \(\lambda\) can be negative would give only +6, and using \(\lambda = \pm 1\) gives \(\pm 2\).

Answer: D

\(\vec{a} + 2\vec{b} + 5\vec{c}\) equals \(\vec{c}\).

Write \(\vec{a} + 2\vec{b} = \lambda\vec{c}\) and \(\vec{b} + 2\vec{c} = \mu\vec{a}\). Substituting \(\vec{a} = \lambda\vec{c} - 2\vec{b}\) into the second:

\[ \begin{aligned} \vec{b} + 2\vec{c} &= \mu\lambda\vec{c} - 2\mu\vec{b} \\ (1 + 2\mu)\vec{b} &= (\mu\lambda - 2)\vec{c} \end{aligned} \]

Since \(\vec{b}\) and \(\vec{c}\) are not collinear, both coefficients vanish: \(\mu = -\frac{1}{2}\) and \(\lambda = -4\). So \(\vec{a} + 2\vec{b} = -4\vec{c}\), and \(\vec{a} + 2\vec{b} + 5\vec{c} = \vec{c}\). The zero vector would be \(\vec{a} + 2\vec{b} + 4\vec{c}\).