Three Dimensional Geometry Questions and Answers

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 \(|\vec{b}_{1}||\vec{b}_{2}| = 21\).

Answer: B

The line is \(\frac{x+3}{-2} = \frac{y-4}{3} = \frac{z-11}{2}\).

Through \((x_{1}, y_{1}, z_{1}) = (-3, 4, 11)\), the direction is the difference of the two points:

\[ \begin{aligned} (1 - (-3),\ -2 - 4,\ 7 - 11) &= (4, -6, -4) \\ &= -2(-2, 3, 2) \end{aligned} \]

Any multiple of a direction vector is a valid set of direction ratios, so \((-2, 3, 2)\) works, giving the second option. Check: the point \((1, -2, 7)\) gives \(\frac{4}{-2} = \frac{-6}{3} = \frac{-4}{2} = -2\).

Options with \(y + 4\) and \(z + 11\) use the point with the wrong signs. The first option's ratios \((2, 3, 4)\) and the last option's \((-2, -3, -4)\) are not proportional to \((4, -6, -4)\).

Answer: E

\(k = -4\).

Two lines through points \(A\) and \(B\) with directions \(\vec{b}_{1}, \vec{b}_{2}\) are coplanar when \(\overrightarrow{AB}\), \(\vec{b}_{1}\) and \(\vec{b}_{2}\) are coplanar, that is their scalar triple product is zero. Here \(A = (1, -1, -10)\), \(B = (4, -3, -1)\), so \(\overrightarrow{AB} = (3, -2, 9)\):

\[ \begin{aligned} \begin{vmatrix} 3 & -2 & 9 \\ 2 & -3 & 8 \\ 1 & k & 7 \end{vmatrix} &= 0 \\ 3(-21 - 8k) + 2(14 - 8) + 9(2k + 3) &= 0 \\ -24 - 6k &= 0, \quad k = -4 \end{aligned} \]

The positive value 4 is the sign slip in the last line; 0, 2 and \(-2\) leave the determinant nonzero, so those lines would be skew.

Answer: D

The point is (9,12,-17).

A general point of the line is \((1 + 4t,\ 2 + 5t,\ -3 - 7t)\). The \(x\)-coordinate 9 needs \(t = 2\), and then:

\[ \begin{aligned} y &= 2 + 5(2) = 12 \\ z &= -3 - 7(2) = -17 \end{aligned} \]

So \((9, 12, -17)\) is on the line. (9,12,-15) has the right \(x\) and \(y\) but the wrong \(z\); (9,15,12) and (12,9,-17) do not come from a single value of \(t\); (-9,-12,17) would need \(t = -2.5\) from \(x\) but \(t = -2.8\) from \(y\).

Answer: B

\(\cos\beta = \frac{3}{7}\).

The direction ratios of the line are the differences of the coordinates:

\[ \begin{aligned} (3 - 1,\ 2 + 1,\ 8 - 2) &= (2, 3, 6) \\ \sqrt{2^{2} + 3^{2} + 6^{2}} &= \sqrt{49} = 7 \end{aligned} \]

The direction cosines are \(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\). The angle with the y-axis uses the second one, so \(\cos\beta = \frac{3}{7}\). The option \(\frac{2}{7}\) is the cosine of the angle with the x-axis.

Answer: A

The unit vector is \(\frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}\).

The direction of the line is the coefficient of the parameter \(s\) in each component:

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

Dividing by the magnitude 3 gives the unit vector. Note the minus sign outside the bracket in the i-component: \(-(5 + 4s)\) contributes \(-4s\). The other options divide the same vector by 4, 5, 6 or 7, which do not give length 1.

Answer: E

The shortest distance is \(10\sqrt{2}\).

The foot of the perpendicular from \((x, y, z)\) to the z-axis is \((0, 0, z)\), so the distance depends only on \(x\) and \(y\):

\[ d = \sqrt{x^{2} + y^{2}} = \sqrt{100 + 100} = 10\sqrt{2} \]

The option \(10\sqrt{3}\) is the distance from the origin, which wrongly includes the z-coordinate, and 10 is the distance to a coordinate plane rather than to the axis.

Answer: B

\(\alpha = 2\).

Two lines are perpendicular when the dot product of their direction ratios is zero.

\[ \begin{aligned} \vec{d_{1}} &= (5 - 3, -1 - 2, -2 - 3) = (2, -3, -5) \\ \vec{d_{2}} &= (\alpha - 1, \alpha - 3, \alpha - 1) \end{aligned} \]

\[ \begin{aligned} 2(\alpha - 1) - 3(\alpha - 3) - 5(\alpha - 1) &= 0 \\ -6\alpha + 12 &= 0 \\ \alpha &= 2 \end{aligned} \]

Check: with \(\alpha = 2\), \(\vec{d_{2}} = (1, -1, 1)\) and \(2 + 3 - 5 = 0\).

Answer: E

a + b = 0.

Rewrite the line in symmetric form by dividing out the coefficients of x and y:

\[ \frac{x + 2}{b/3} = \frac{y - 1}{3/2} = \frac{z - 3}{1} \]

The line passes through (-2, 1, 3), so a = -2. Its direction \(\left(\frac{b}{3}, \frac{3}{2}, 1\right)\) must match \(\left(\frac{2}{3}, \frac{3}{2}, 1\right)\), and since the last two components already agree, \(\frac{b}{3} = \frac{2}{3}\) and b = 2.

So a + b = -2 + 2 = 0. Reading a as +2 from '3x + 6' (sign slip) gives 4, and reading b as \(\frac{2}{3}\) without the factor 3 gives a non-integer.

Answer: A

The key gives \(\cos^{-1}\left(\frac{-4}{21}\right)\), the angle between the two direction vectors as given.

\[ \begin{aligned} \cos\theta &= \frac{2(3) + (-1)(6) + 2(-2)}{\sqrt{9}\sqrt{49}} \\ &= \frac{6 - 6 - 4}{3 \times 7} = \frac{-4}{21} \end{aligned} \]

The angle between two lines is often quoted as the acute one, \(\cos^{-1}\frac{4}{21}\), but that value is not among the options; the options list only the angle between the direction vectors (2, -1, 2) and (3, 6, -2), which is obtuse. \(\frac{-4}{7}\) leaves out the factor 3 for \(|(2, -1, 2)|\), and the positive fractions come from errors in the dot product.