Three Dimensional Geometry Questions and Answers

Answer: C

The only possible value is \(\lambda = -1\).

Line 1 is the set of points \((t, t, 1)\). The foot of the perpendicular from \(P(\lambda, \lambda, \lambda)\) minimises \(2(\lambda - t)^{2} + (\lambda - 1)^{2}\), so \(t = \lambda\) and \(Q = (\lambda, \lambda, 1)\).

Line 2 is the set of points \((s, -s, -1)\). Minimising \((\lambda - s)^{2} + (\lambda + s)^{2} + (\lambda + 1)^{2}\) gives \(s = 0\), so \(R = (0, 0, -1)\).

\[ \begin{aligned} \vec{PQ} &= (0, 0, 1 - \lambda), \quad \vec{PR} = (-\lambda, -\lambda, -1 - \lambda) \\ \vec{PQ} \cdot \vec{PR} &= -(1 - \lambda)(1 + \lambda) = \lambda^{2} - 1 = 0 \end{aligned} \]

So \(\lambda = \pm 1\). But for \(\lambda = 1\), P = (1, 1, 1) lies on line 1 itself, so Q = P and there is no angle QPR. Hence \(\lambda = -1\). The values \(\pm\sqrt{2}\) do not satisfy \(\lambda^{2} = 1\).

Answer: B

The value is \(\frac{8}{9}\).

Direction cosines satisfy \(\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1\). Square the given sum:

\[ \begin{aligned} \left(\tfrac{5}{3}\right)^{2} &= 1 + 2S \\ \tfrac{25}{9} - 1 &= 2S \\ S &= \tfrac{16}{18} = \tfrac{8}{9} \end{aligned} \]

where \(S\) is the required sum of products. Forgetting to halve gives \(\frac{16}{9}\) (not offered); \(\frac{11}{9}\) and \(\frac{7}{9}\) come from using 2 or 3 for the sum of squares.

Answer: C

\(Q = \) (3,1,5).

A general point of the given line is \(Q = (1 + 2t,\ t,\ 2 + 3t)\). With \(P = (6, 1, 3)\), \(\vec{PQ} = (2t - 5,\ t - 1,\ 3t - 1)\), which must be perpendicular to the direction (2, 1, 3):

\[ \begin{aligned} 2(2t - 5) + (t - 1) + 3(3t - 1) &= 0 \\ 14t - 14 &= 0 \Rightarrow t = 1 \end{aligned} \]

So \(Q = (3, 1, 5)\). The other points do not lie on the given line: for (2,1,3), \(\frac{2 - 1}{2} = \frac{1}{2}\) but \(\frac{y}{1} = 1\).

Answer: D

The angle is \(\cos^{-1}\left(\frac{1}{3}\right)\).

The direction ratios are (1, -1, -1) and (2, 2, -2):

\[ \begin{aligned} \cos\theta &= \frac{|2 - 2 + 2|}{\sqrt{3}\,\sqrt{12}} \\ &= \frac{2}{6} = \frac{1}{3} \end{aligned} \]

The dot product is 2 and the product of the lengths is \(\sqrt{3} \times 2\sqrt{3} = 6\), so only \(\frac{1}{3}\) fits. The surd options come from slips in the lengths, such as taking \(|(2, 2, -2)|\) as \(\sqrt{8}\) or \(\sqrt{6}\).

Answer: E

A unit vector is \(\frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} - 2\hat{k})\).

The direction is the difference of the two points:

\[ \begin{aligned} \vec{d} &= (13 - 10,\ 9 - 8,\ 4 - 6) = (3, 1, -2) \\ |\vec{d}| &= \sqrt{9 + 1 + 4} = \sqrt{14} \end{aligned} \]

Dividing by \(\sqrt{14}\) gives the answer. The option with \(+2\hat{k}\) has the sign of the \(z\)-change wrong, and the \(\sqrt{17}\) options are not unit vectors (\(9 + 1 + 4 = 14\), not 17).

Answer: A

The line passes through (-3,0,-3).

A point lies on the line when all three ratios are equal. For \((-3, 0, -3)\):

\[ \frac{-3 + 1}{2} = \frac{0 - 4}{4} = \frac{-3 - 2}{5} = -1 \]

All equal \(-1\), so the point is on the line (it is the point with parameter \(-1\)). For \((3, 0, 5)\) the ratios are \(2, -1, \frac{3}{5}\); for \((-3, 0, 5)\), \(-1, -1, \frac{3}{5}\); for \((3, 4, 5)\), \(2, 0, \frac{3}{5}\); none is a single value.

Answer: B

The values are 1, -4.

Two lines are coplanar when the vector joining a point on each and their two direction vectors are coplanar, so their scalar triple product is zero. Points \((4, 3, -2)\) and \((3, 4, -3)\) give the joining vector \((-1, 1, -1)\):

\[ \begin{vmatrix} -1 & 1 & -1 \\ m & 2 & 1 \\ 1 & 1 & m \end{vmatrix} = -m^{2} - 3m + 4 = 0 \]

so \(m^{2} + 3m - 4 = (m + 4)(m - 1) = 0\), giving \(m = 1\) or \(-4\). The other pairs fail: for example \(m = 4\) makes the determinant \(-24\).

Answer: D

\(\alpha =\) \(-3\).

For \((-1, -2, -3)\) the first two ratios are \(\frac{0}{4} = 0\) and \(\frac{0}{-3} = 0\), so the third must also be zero:

\[ \frac{-3 - \alpha}{-2} = 0 \;\Rightarrow\; \alpha = -3 \]

In fact the point is the line's own base point \((-1, -2, \alpha)\). \(\alpha = 3\) changes the sign, and the other values leave the third ratio non-zero.

A useful habit: when a point's first coordinates already match the line's base point, the parameter is 0, and every remaining coordinate must match too.

Answer: A

The distance is \(\frac{\sqrt{2}}{2}\).

The line passes through A with position vector \(\hat{j}\) in the direction \(\vec{d} = \hat{i} + \hat{j}\). The distance of P from it is \(\frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}\).

\[ \begin{aligned} \vec{AP} &= 2\hat{j} - \hat{j} = \hat{j} \\ \hat{j} \times (\hat{i} + \hat{j}) &= -\hat{k} \\ d &= \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \end{aligned} \]

A quick check in the plane: the line is y = x + 1, and the distance of (0, 2) from x - y + 1 = 0 is \(\frac{|0 - 2 + 1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}\). The other values come from dividing by \(\sqrt{3}\) or 3 instead of \(\sqrt{2}\).

Answer: A

The Cartesian form is \(\frac{x - 2}{3} = \frac{y + 7}{7} = \frac{z - 11}{-13}\).

A line through \((x_{1}, y_{1}, z_{1})\) with direction ratios a, b, c is \(\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}\). Here the point is (2, -7, 11) and the direction ratios are 3, 7, -13.

The y-term is \(y - (-7) = y + 7\). Options writing x + 2 or y - 7 reverse the signs of the point's coordinates, the option with 13 in the z-denominator loses the minus sign of the direction ratio, and the last option uses neither the point nor the direction.