Three Dimensional Geometry Questions and Answers

Answer: B

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

From \(\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\), the line passes through the point \((2, -1, 3)\) and has direction ratios \((3, 2, 2)\). The vector form is \(\vec{r} = \vec{a} + \mu\vec{b}\) with \(\vec{a}\) the point and \(\vec{b}\) the direction.

Option A gets the signs of the point and the direction wrong. Options C, D and E swap the roles of the point and the direction, or alter their components.

Answer: E

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

The direction of each line is the coefficient of its parameter: (1, 2, 2) for the first and (3, 2, 6) for the second.

\[ \begin{aligned} \cos\theta &= \frac{1(3) + 2(2) + 2(6)}{\sqrt{1 + 4 + 4}\,\sqrt{9 + 4 + 36}} \\ &= \frac{19}{3 \times 7} = \frac{19}{21} \end{aligned} \]

\(\frac{19}{12}\) is greater than 1 and cannot be a cosine. \(\frac{17}{21}\) and \(\frac{9}{21}\) come from slips in the dot product, and \(2\cos^{-1}\) doubles the angle.

Answer: B

The line is \(\frac{x - 1}{3} = \frac{y - 1}{2} = \frac{z - 1}{1}\).

First put the given line in standard form. \(\frac{2 - x}{-3} = \frac{x - 2}{3}\), so the line is \(\frac{x - 2}{3} = \frac{y - 2}{2} = \frac{z - 4}{1}\), with direction ratios 3, 2, 1.

A parallel line through (1, 1, 1) has the same direction ratios, giving the keyed equation. Taking the -3 at face value gives direction (-3, 2, 1), which ignores the sign of \(2 - x\). The options with \(z + 1\) pass through \(z = -1\), not 1.

Answer: E

The shortest distance is 2.

The second line is \((\hat{i} - \hat{j} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})\), so both lines have direction \(\vec{d} = (2, 2, 1)\) and are parallel. Use a point on each: (1, 1, 3) and (1, -1, 1).

\[ \begin{aligned} \vec{w} &= (1, -1, 1) - (1, 1, 3) = (0, -2, -2) \\ \vec{w} \times \vec{d} &= (2, -4, 4),\ |\vec{w} \times \vec{d}| = 6 \\ d &= \frac{6}{|\vec{d}|} = \frac{6}{3} = 2 \end{aligned} \]

6 is the size of the cross product before dividing by \(|\vec{d}| = 3\). The skew-line formula does not apply here because the lines are parallel.

Answer: E

The angle is \(\cos^{-1}\left(\frac{1}{\sqrt{7}}\right)\).

Use the direction ratios \((-4, 3, 5)\) and \((2, 1, 3)\):

\[ \begin{aligned} \cos\theta &= \frac{-8 + 3 + 15}{\sqrt{50}\,\sqrt{14}} = \frac{10}{\sqrt{700}} \\ &= \frac{10}{10\sqrt{7}} = \frac{1}{\sqrt{7}} \end{aligned} \]

The options with \(3\sqrt{7}\) or \(2\sqrt{7}\) come from errors in the magnitudes, \(\frac{2}{7}\) squares part of the result, and \(\sqrt{2}\) exceeds 1, so it cannot be a cosine.

Answer: E

The vector form is \(\vec{r} = (2 + \mu)\hat{i} + (1 - \mu)\hat{j} + (1 - 2\mu)\hat{k}\).

Set each ratio equal to \(\mu\): \(x - 2 = \mu\), \(y - 1 = -\mu\), \(z - 1 = -2\mu\). So x = 2 + μ, y = 1 - μ, z = 1 - 2μ: the point (2, 1, 1) plus μ times the direction (1, -1, -2).

Each other option gets one component wrong: \(1 + 2\mu\) and \(1 + \mu\) flip a sign, \(1 - 2\mu\) in the y-component uses the z-ratio, and \(2 + 3\mu\) uses a direction ratio of 3 that is not in the line.

Answer: D

The Gaussian curvature is \(\det\begin{pmatrix} L & M \\ M & N \end{pmatrix}\) / \(\det\begin{pmatrix} E & F \\ F & G \end{pmatrix}\).

Gaussian curvature is the determinant of the shape operator, which is the second fundamental form expressed relative to the first: its matrix is \(\mathcal{F}_{I}^{-1}\mathcal{F}_{II}\). Taking determinants,

\[ K = \frac{LN - M^{2}}{EG - F^{2}} \]

The determinant of the first form alone is the squared area element, and that of the second alone depends on the parametrisation; their product has no geometric meaning. Only the ratio is independent of the parametrisation.

Answer: A

\(\bar{t}'\) is orthogonal to \(\bar{t}\) and \(\bar{b}\).

The first Frenet-Serret formula for a unit speed curve is \(\bar{t}' = \kappa\bar{n}\). So \(\bar{t}'\) is a multiple of the normal, and the normal is perpendicular to both the tangent and the binormal.

The orthogonality to \(\bar{t}\) can also be seen directly: \(\bar{t} \cdot \bar{t} = 1\), and differentiating gives \(2\bar{t} \cdot \bar{t}' = 0\).

\(\bar{t}'\) is parallel to \(\bar{n}\), so it is not orthogonal to \(\bar{n}\) (unless \(\kappa = 0\)), which rules out B and C, and D is wrong because it is orthogonal to \(\bar{t}\) and \(\bar{b}\).

Answer: A

The Gaussian curvature is always \(\lt 0\).

For a graph \(z = f(x, y)\),

\[ K = \frac{f_{xx}f_{yy} - f_{xy}^2}{(1 + f_x^2 + f_y^2)^2} \]

Here \(f = x^{2}/2 - y^{2}/3\), so \(f_{xx} = 1\), \(f_{yy} = -2/3\) and \(f_{xy} = 0\). The numerator is \(-2/3\) at every point, and the denominator is positive, so \(K \lt 0\) everywhere.

Every point of a hyperbolic paraboloid is a saddle (hyperbolic) point, so the curvature is never 0 or positive, ruling out the other options.

Answer: C

\(X^{-1}Y\) may not be symmetric, but has real eigenvalues.

\(X^{-1}Y\) is the matrix of the shape (Weingarten) operator in the basis \(\{\sigma_{u}, \sigma_{v}\}\). That basis need not be orthonormal, so the matrix need not be symmetric. But the shape operator is self-adjoint with respect to the first fundamental form, so its eigenvalues, the principal curvatures, are real.

It need not be non-singular: its determinant is the Gaussian curvature \(K = \det Y / \det X\), which is 0 on a cylinder or plane. So A and D fail, and B fails because the matrix is generally not symmetric.