Three Dimensional Geometry: questions with answers

The identity is $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. From it, $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 3 - 1 = 2$ and $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2 \cdot 1 - 3 = -1$, so those are true. The plain sum $\cos\alpha + \cos\beta + \cos\gamma = 1$ is not an identity: for the $x$-axis itself it is $1 + 0 + 0 = 1$, but for a line making $60^\circ$ with each axis it is $\frac32$.

(R) is the identity satisfied by the direction cosines of every line, so it is true. (A) is true and (R) is exactly why: the squares of the direction cosines of every line add to $1$, and three angles of $60^\circ$ give only $\frac34$. So both are true and (R) is the correct explanation of (A). Neither the plain sum of cosines nor 'any three numbers will do' has a place here.

Parallel lines need ALL THREE quotients $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ equal. Here the first two are $\frac12$ but the third is $\frac{2}{-4}$, so the direction ratios are not proportional and the lines are not parallel. Checking one or two pairs is not enough.

The components $-18, 12, -4$ are direction RATIOS -- numbers proportional to the cosines, not the cosines themselves. Dividing by $\sqrt{ 324 + 144 + 16 = 484 } = 22$ gives the direction cosines $\langle -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \rangle$, whose squares add to $1$. Dividing by the sum of the squares or by the sum of the moduli does not give a unit triple.

Direction cosines already form unit vectors ($l^2 + m^2 + n^2 = 1$ for each line), so no division is needed and the sum of products IS the cosine -- but of the angle between the two directed lines, which may be obtuse. The angle between two LINES is taken acute, so the modulus is part of the formula: $\cos\theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$. Dropping it reports $\theta > 90^\circ$ whenever the sum is negative.

The line's direction is $\vec b = 3\hat i - \hat j$, and a vector is perpendicular to the line exactly when its dot product with $\vec b$ is ZERO: $(3)(1) + (-1)(3) + (0)(5) = 0$. $9\hat i - 3\hat j$ is $3\vec b$ -- a vector PARALLEL to the line, pointing along it rather than across it, and proportionality is the parallel condition. $5\hat i + \hat j + 6\hat k$ and $2\hat i - 2\hat j$ give $14$ and $8$, neither of them zero.

The line through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$: the coordinates are SUBTRACTED, so a negative coordinate appears as a plus, and a zero direction ratio stays as a zero denominator. So $\frac{x-2}{0} = \frac{y+1}{1} = \frac{z-3}{1}$. Copying the coordinates' signs gives $\frac{x+2}{0} = \frac{y-1}{1} = \frac{z+3}{1}$, and replacing the zero by $1$ changes the direction.

Lines 1 and 2 are correct. Line 3 is the first wrong line: the foot lies ON the mirror line; the image lies on the far side, with $M$ as the midpoint of $PP'$, so $P' = 2M - P = (0, 5, 1)$.

Direction ratios of $AB$: $\langle 2, 3, 4 \rangle$; of $BC$: $\langle 4, 6, 8 \rangle$. These are PROPORTIONAL (every ratio of $BC$ is $2$ times the corresponding ratio of $AB$), so $AB \parallel BC$; since $B$ lies on both, $A$, $B$, $C$ are collinear. The ratios need not be equal, and adding coordinates ($\langle 4, 7, 10 \rangle$, $\langle 10, 16, 22 \rangle$) tests nothing about the segments.

The ratios $-3, 1, 5$ and $-1, 2, 5$ are not proportional, so the lines are not parallel -- but non-parallel lines in space may still miss each other. The test is $(\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2)$, which is the shortest distance times $|\vec b_1 \times \vec b_2|$: with $\vec b_1 \times \vec b_2 = -5\hat i + 10\hat j - 5\hat k$ and $\vec a_2 - \vec a_1 = 2\hat i + \hat j$ it is $(2)(-5) + (1)(10) + (0)(-5) = 0$. Zero means the gap is zero, so the lines are coplanar and, not being parallel, they meet at one point.

Line 1 is correct. Line 2 is the first wrong line: $\vec a_1$ and $\vec a_2$ are arbitrary points of the two lines -- slide either line along itself and that length changes, while the gap between the lines does not. The shortest distance is the part of $\vec a_2 - \vec a_1$ along the common perpendicular $\vec b_1 \times \vec b_2 = 8\hat i + 8\hat j + 4\hat k$: $d = 9$, which is smaller. Line 3 repeats Line 2.

Lines 1 and 2 are both correct -- that triple really does satisfy the first condition. Line 3 is the first wrong line: one condition is satisfied by a whole PLANE of directions, so satisfying it proves nothing until the second is tested, and there this triple gives $(2)(0) + (-1)(-3) + (0)(2)$, not zero. Solving the two conditions together gives $13 : -2 : -3$.

Line 1 is correct. Line 2 is the first wrong line: $\sqrt{ 6^2 + (-1)^2 + 8^2 }$ is the distance of $P$ from the ORIGIN. The distance from the axis is $PM = \sqrt{ 6^2 + 8^2 } = 10$, which omits the coordinate along the axis. Line 3 follows from Line 2.

Line 1 is correct. Line 2 is the first wrong line: the distance from a line is the length of the PERPENDICULAR, $PM = \sqrt{ 4 + 9 + 36 } = 7$; $A$ is just one point of the line, and $PA$ is a slant length. Line 3 follows from Line 2.

'Measured parallel to the $z$-axis' means: travel from $P$ along the line through $P$ parallel to that axis until the given line is met. Only the other two coordinates have to match, which happens at $\lambda = 1$, at the point $(3, 3, 4)$; the distance is the change in the $z$-coordinate, $1$. The PERPENDICULAR distance from $P$ to the line, $\sqrt{\frac{5}{14}}$, is a different (shorter) length.