Three Dimensional Geometry Questions and Answers

Answer: B

The line meets the zx-plane at (19,0,5).

The zx-plane is y = 0. Set each ratio equal to \(\lambda\) and put y = 0.

\[ \begin{aligned} \frac{0 - 2}{-1} &= 2 = \lambda \\ x &= -3 + 11(2) = 19 \\ z &= -1 + 3(2) = 5 \end{aligned} \]

(19, 2, 5) keeps y = 2, so it is not on the plane. (-3, 2, 0) and (0, 2, -1) mix up the coordinates of the base point (-3, 2, -1) and do not lie on the line.

Answer: A

The line is \(\vec{r} = \hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})\).

Take the point (0, 0, 1), with position vector \(\hat{k}\), and the direction from it to (1, 1, 0):

\[ \begin{aligned} (1, 1, 0) - (0, 0, 1) &= (1, 1, -1) \\ &= \hat{i} + \hat{j} - \hat{k} \end{aligned} \]

Check: \(\lambda = 1\) gives (1, 1, 0). The other options either start from a point not on the line (\(\hat{j}\), \(\hat{i}\), \(\hat{i} + \hat{j} + \hat{k}\)) or use a direction not parallel to (1, 1, −1).

Answer: A

A parallel vector is \(-3\hat{i} + 5\hat{j} - 2\hat{k}\).

In \(\vec{r} = \vec{a} + \lambda\vec{b}\), the line is parallel to \(\vec{b} = 3\hat{i} - 5\hat{j} + 2\hat{k}\), and any non-zero multiple of \(\vec{b}\) is parallel too. Option A is \(-\vec{b}\).

\(3\hat{i} + 5\hat{j} + 2\hat{k}\) changes only one sign, so it is not a multiple. \(\hat{i} - 11\hat{j} + 101\hat{k}\) is the position vector of a point on the line, not its direction, and the other two are not multiples of \(\vec{b}\).

Answer: A

The line is \(\frac{x - 1}{1} = \frac{y + 1}{1} = \frac{z}{2}\).

A general point of the given line is \((1 + t, -1 + t, 1 - t)\). The direction from (1, −1, 0) to it is \((t, t, 1 - t)\), and this must be perpendicular to the line's direction (1, 1, −1):

\[ t + t - (1 - t) = 0 \Rightarrow t = \frac{1}{3} \]

The direction is then \(\left(\frac{1}{3}, \frac{1}{3}, \frac{2}{3}\right) \propto (1, 1, 2)\), giving option A. Option E has direction (1, 1, −2), which is not perpendicular to (1, 1, −1), and options B, C and D do not pass through (1, −1, 0).

Answer: E

The value of k is 13.

The line through (1, 2, 3) and (3, 4, k) has direction ratios \((3 - 1, 4 - 2, k - 3) = (2, 2, k - 3)\). These must be proportional to the given denominators (1, 1, 5):

\[ \begin{aligned} \frac{2}{1} = \frac{2}{1} &= \frac{k - 3}{5} \\ k - 3 &= 10 \Rightarrow k = 13 \end{aligned} \]

The given equation already passes through (3, 4, k), so only the direction matters. 10 is the value of k - 3, the step before the answer, and the smaller values do not make the direction ratios proportional.

Answer: B

The length is \(5\sqrt{5}\).

The projections of a segment on the three axes are the differences of the coordinates of its ends, so the length is the square root of the sum of their squares:

\[ \begin{aligned} L &= \sqrt{5^{2} + 6^{2} + 8^{2}} = \sqrt{125} \\ &= 5\sqrt{5} \end{aligned} \]

Adding the projections (19) or picking one of them (5 or 6) ignores that they are perpendicular components. \(6\sqrt{5}\) is \(\sqrt{180}\) and \(6\sqrt{6}\) is \(\sqrt{216}\), neither of which is the sum of the squares.

Answer: C

The shortest distance is 3.

A general point of the line is \(Q = (4 - 2t, 4 + 2t, 6 + t)\), with direction (-2, 2, 1). For the foot of the perpendicular from P(2, 3, 4), \(\vec{PQ}\) must be perpendicular to the direction:

\[ \begin{aligned} \vec{PQ} &= (2 - 2t, 1 + 2t, 2 + t) \\ \vec{PQ} \cdot (-2, 2, 1) &= 9t = 0 \Rightarrow t = 0 \end{aligned} \]

So the foot is (4, 4, 6) and the distance is \(\sqrt{2^{2} + 1^{2} + 2^{2}} = \sqrt{9} = 3\). The other values do not come from a perpendicular: 9 is the square of the distance, and \(\sqrt{5}\) and \(\sqrt{3}\) correspond to no point of the line at right angles to P.

Answer: D

P is (7, 10, 13).

The direction from A(1, 2, 3) to B(4, 6, 8) is (3, 4, 5), so points of the line are \((1 + 3t, 2 + 4t, 3 + 5t)\). For x = 7, \(1 + 3t = 7\) gives t = 2:

\[ (1 + 6, 2 + 8, 3 + 10) = (7, 10, 13) \]

So P lies beyond B, with AP = 2AB. The options with -10 or -13 have the wrong signs, and (7, 10, 12) and (7, 10, 15) have the right y but a z that does not lie on the line.

Answer: C

k is 9.

Put x = 3 and y = 6 into the equation of the line:

\[ \frac{3 - 1}{1} = \frac{6 - 2}{2} = 2 = \frac{k - 3}{3} \]

The first two fractions agree, so the point can lie on the line, and \(k - 3 = 6\) gives k = 9. 2 is the common value of the fractions, not k, and 3 is the z-coordinate of the base point (1, 2, 3).

Answer: C

The angle is \(\cos^{-1}\left(\frac{19}{21}\right)\).

The angle between two lines is the angle between their direction vectors, here \(\vec{b}_{1} = \hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b}_{2} = 3\hat{i} + 2\hat{j} + 6\hat{k}\). The position vectors play no part.

\[ \begin{aligned} \vec{b}_{1} \cdot \vec{b}_{2} &= 3 + 4 + 12 = 19 \\ |\vec{b}_{1}| &= \sqrt{1 + 4 + 4} = 3 \\ |\vec{b}_{2}| &= \sqrt{9 + 4 + 36} = 7 \\ \cos\theta &= \frac{19}{3 \times 7} = \frac{19}{21} \end{aligned} \]

The position vectors \(3\hat{i} + 2\hat{j} - 4\hat{k}\) and \(5\hat{i} - 2\hat{j}\) only fix where each line sits, not its direction. The other options have denominators 13, 19 and 17, which cannot arise, since \(|\vec{b}_{1}||\vec{b}_{2}| = 21\).