Corrected exercises: dot product, projections and planes (MATH 133)

a) Multiply coordinate by coordinate and add: $\mathbf{u} \cdot \mathbf{v} = 2 \times 1 + 1 \times 1 + (-2) \times 0 = 3$. The lengths are $\|\mathbf{u}\| = \sqrt{4 + 1 + 4} = 3$ and $\|\mathbf{v}\| = \sqrt{1 + 1 + 0} = \sqrt{2}$. Then $\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|} = \dfrac{3}{3\sqrt{2}} = \dfrac{1}{\sqrt{2}}$, so $\theta = \dfrac{\pi}{4}$. Two slips cost the mark here: forgetting the square root in a length (writing $\|\mathbf{u}\| = 9$), and dividing by only one of the two lengths. The values in a MATH 133 exam are chosen so that $\cos\theta$ lands on $0$, $\pm\frac{1}{2}$, $\pm\frac{1}{\sqrt{2}}$ or $\pm\frac{\sqrt{3}}{2}$; if yours does not, recheck a length before reaching for $\arccos$.

b) $\mathbf{u} \cdot \mathbf{w} = 2 \times 1 + 1 \times 0 + (-2) \times 1 = 0$. Since neither vector is zero, $\cos\theta = 0$ and $\theta = \frac{\pi}{2}$: they are orthogonal. The sentence must mention that both vectors are non-zero. The zero vector has dot product $0$ with everything, and by convention it is called orthogonal to every vector, but it has no direction and no angle.

c) Orthogonal means the dot product is zero, and here it is a quadratic in $k$: $(k, k, 1) \cdot (k, -3, 2) = k^2 - 3k + 2 = (k - 1)(k - 2)$. So $k = 1$ or $k = 2$. Check both, because a sign slip in the middle coordinate would change the roots: $(1, 1, 1) \cdot (1, -3, 2) = 1 - 3 + 2 = 0$ and $(2, 2, 1) \cdot (2, -3, 2) = 4 - 6 + 2 = 0$. Answering only $k = 1$ loses half the part: the question says every real $k$.

d) $\mathbf{u} \cdot \mathbf{x} = -2 + 2 - 2 = -2$. Lengths are positive, so $\cos\theta$ has the SIGN of the dot product: positive means acute, zero means right, negative means obtuse. Here the angle is obtuse, and computing it would be wasted time. This sign test is exactly what the figure shows, and it is the fastest way to answer questions of the type is the angle at vertex $A$ of this triangle obtuse.

e) $\mathbf{u} \cdot (-\mathbf{v}) = -3$ while the lengths are unchanged, so $\cos\theta' = -\dfrac{1}{\sqrt{2}}$ and $\theta' = \dfrac{3\pi}{4}$. Reversing one vector replaces $\theta$ by $\pi - \theta$. Keep this in mind for the rest of the chapter: a line or a plane does not care about the sign of its direction or normal vector, so an angle between LINES or between PLANES is always reported as the acute one, whereas an angle between two VECTORS can be anything in $[0, \pi]$.

a) $\mathbf{u} \cdot \mathbf{v} = 3 + 1 + 2 = 6$ and $\|\mathbf{v}\|^2 = 3$, so $\mathbf{p} = \dfrac{6}{3}(1, 1, 1) = (2, 2, 2)$. Note that $\|\mathbf{v}\|^2$ is computed WITHOUT a square root: $\|\mathbf{v}\|^2 = \mathbf{v} \cdot \mathbf{v}$. The most frequent wrong line on this formula divides by $\|\mathbf{v}\|$ instead of $\|\mathbf{v}\|^2$ and writes $\frac{6}{\sqrt{3}}(1, 1, 1)$, a vector $\sqrt{3}$ times too long. Part e) shows why that version cannot be right.

b) $\mathbf{u} - \mathbf{p} = (3, 1, 2) - (2, 2, 2) = (1, -1, 0)$, and $(1, -1, 0) \cdot (1, 1, 1) = 1 - 1 + 0 = 0$: orthogonal, as it must be. So $\mathbf{u} = (2, 2, 2) + (1, -1, 0)$, the first vector a multiple of $\mathbf{v}$, the second orthogonal to it. This check costs five seconds and catches almost every arithmetic slip in a): if $\mathbf{u} - \mathbf{p}$ does not dot to zero with $\mathbf{v}$, the projection is wrong. It is also the picture on the figure: the leftover arrow stands at a right angle on the line of $\mathbf{v}$.

c) The scalar component is $\dfrac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|} = \dfrac{6}{\sqrt{3}} = 2\sqrt{3}$, a signed NUMBER, whereas $\mathbf{p}$ is a VECTOR. Here $\|\mathbf{p}\| = \sqrt{4 + 4 + 4} = 2\sqrt{3}$, the same value, because $\mathbf{u} \cdot \mathbf{v} > 0$. Had the dot product been negative, the scalar component would be $-\|\mathbf{p}\|$ and $\mathbf{p}$ would point against $\mathbf{v}$. Exam questions say which one they want, projection (a vector) or scalar component (a number), and giving the other loses the part.

d) $\operatorname{proj}_{\mathbf{u}}\mathbf{v} = \dfrac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2}\,\mathbf{u} = \dfrac{6}{14}(3, 1, 2) = \left(\dfrac{9}{7}, \dfrac{3}{7}, \dfrac{6}{7}\right)$. It is NOT $\mathbf{p}$: it lies along $\mathbf{u}$, not along $\mathbf{v}$, and has a different length, $\dfrac{6}{\sqrt{14}}$ instead of $2\sqrt{3}$. The dot product in the numerator is symmetric, the rest of the formula is not. Write the subscript, the vector you project ON, before writing anything else.

e) With $2\mathbf{v} = (2, 2, 2)$: $\dfrac{\mathbf{u} \cdot 2\mathbf{v}}{\|2\mathbf{v}\|^2}\,2\mathbf{v} = \dfrac{12}{12}(2, 2, 2) = (2, 2, 2)$. With $-\mathbf{v}$: $\dfrac{-6}{3}(-1, -1, -1) = (2, 2, 2)$. The scale factor appears twice on top and twice below, so it cancels: the projection depends only on the line through the origin spanned by $\mathbf{v}$. This is precisely why the denominator must be $\|\mathbf{v}\|^2$. With $\|\mathbf{v}\|$ alone the answer would double when $\mathbf{v}$ doubles, and a shadow on a line cannot depend on which arrow you used to name that line.

a) $\mathbf{w} = Q - P_0 = (5 - 1, 5 - 0, 3 - 1) = (4, 5, 2)$. Then $\mathbf{w} \cdot \mathbf{d} = 4 + 10 + 4 = 18$ and $\|\mathbf{d}\|^2 = 1 + 4 + 4 = 9$, so $\operatorname{proj}_{\mathbf{d}}\mathbf{w} = \dfrac{18}{9}(1, 2, 2) = (2, 4, 4)$. The ratio $\frac{18}{9} = 2$ is worth keeping: it is the value of the parameter $t$ at the foot of the perpendicular.

b) Walk from $P_0$ along the line by exactly the projection: $F = P_0 + (2, 4, 4) = (3, 4, 5)$, which is $X(2)$, so $F$ is indeed on $L$. The projection answers where on the line; it does not answer how far from the line.

c) The distance is the length of what is LEFT: $\mathbf{w} - \operatorname{proj}_{\mathbf{d}}\mathbf{w} = (4, 5, 2) - (2, 4, 4) = (2, 1, -2)$, of length $\sqrt{4 + 1 + 4} = 3$. This is the trap of the exercise and it costs the whole part: $\|\operatorname{proj}_{\mathbf{d}}\mathbf{w}\| = 6$ is the distance from $P_0$ to $F$ ALONG the line, and it depends on which point $P_0$ was used to write $L$. The distance from $Q$ to the line cannot depend on that choice, and the leftover does not.

d) $\overrightarrow{FQ} = Q - F = (2, 1, -2)$, the leftover of c), and $(2, 1, -2) \cdot (1, 2, 2) = 2 + 2 - 4 = 0$: the segment $FQ$ meets $L$ at a right angle, as on the solution figure. Pythagoras in the right triangle $P_0FQ$: $\|\mathbf{w}\|^2 = 16 + 25 + 4 = 45$, and $6^2 + 3^2 = 36 + 9 = 45$. Both checks hold, and each takes a line.

e) $\overrightarrow{X(t)Q} = \mathbf{w} - t\mathbf{d} = (4 - t, 5 - 2t, 2 - 2t)$, so $\|\overrightarrow{X(t)Q}\|^2 = (4 - t)^2 + (5 - 2t)^2 + (2 - 2t)^2 = 9t^2 - 36t + 45 = 9(t - 2)^2 + 9$. The square is never negative, so the minimum is $9$, reached at $t = 2$: same foot $F = X(2)$, same distance $\sqrt{9} = 3$. No derivative is needed, and none is expected in MATH 133. The two methods are the same computation in disguise: expanding gives $\|\mathbf{w}\|^2 - 2t(\mathbf{w} \cdot \mathbf{d}) + t^2\|\mathbf{d}\|^2$, whose vertex sits at $t = \dfrac{\mathbf{w} \cdot \mathbf{d}}{\|\mathbf{d}\|^2}$, the coefficient of the projection.

a) $\mathbf{n} \cdot (X - P) = 0$ reads $1(x - 2) - 2(y + 1) + 2(z - 3) = 0$, that is $x - 2y + 2z = 2 + 2 + 6 = 10$. The shortcut is to write $x - 2y + 2z = d$ straight from the normal and find $d$ by substituting $P$: $2 - 2(-1) + 2(3) = 10$. Always substitute $P$ back into the final equation; a sign slip on $d$ is the commonest error here and it moves the plane without changing its tilt.

b) For $A$: $\overrightarrow{PA} = (2, 2, 1)$ and $\mathbf{n} \cdot \overrightarrow{PA} = 2 - 4 + 2 = 0$, so $\overrightarrow{PA}$ lies in the plane and $A$ is on it; substitution agrees, $4 - 2 + 8 = 10$. For $B$: $1 - 2 + 10 = 9 \neq 10$, so $B$ is NOT on it. Note that $B$ is off by one unit of the equation, not one unit of distance; the actual distance would be $\frac{1}{\|\mathbf{n}\|} = \frac{1}{3}$, as exercise 6 will show.

c) A plane perpendicular to a line has the DIRECTION of the line as its normal: $\mathbf{n} = (2, -1, 1)$. Through $Q$: $2x - y + z = 2 - 1 + 1 = 2$. The point $(3, 0, -1)$ of the line is irrelevant, since the plane is asked to pass through $Q$, and a copy that uses it instead writes $2x - y + z = 5$, a parallel plane in the wrong place. Perpendicular to a line and containing a line are opposite situations: for the second, the direction would have to be ORTHOGONAL to $\mathbf{n}$.

d) Parallel planes share their normal, so the answer is $2x + y - 2z = d$ with $d = 2(1) + 0 - 2(2) = -2$: the plane $2x + y - 2z = -2$. Only $d$ changes. The given plane is $2x + y - 2z = 5$ and $R$ gives $-2 \neq 5$, so $R$ is not on it and the new plane is a genuine different, parallel plane. Had $R$ satisfied the original equation, the only answer would have been the original plane itself.

e) In $\mathbb{R}^3$ the equation $3x - y = 7$ is $3x - y + 0z = 7$: a PLANE with normal $(3, -1, 0)$, not a line. The missing $z$ means $z$ is free, so the plane is vertical: it contains the whole vertical line through every point of the line $3x - y = 7$ in the $xy$-plane. For instance $(3, 2, 0)$ and $(3, 2, 5)$ both satisfy it, since $9 - 2 = 7$. Calling it a line is the classic error of carrying the picture of $\mathbb{R}^2$ into $\mathbb{R}^3$: one linear equation in three unknowns always leaves two free parameters.

a) First the test: $\mathbf{d} \cdot \mathbf{n} = 2 - 1 + 3 = 4 \neq 0$, so $L_1$ is not parallel to $\Pi$ and meets it in exactly one point. Substitute $X(t) = (t, 1 + t, 1 + t)$ into the equation: $2t - (1 + t) + 3(1 + t) = 4t + 2 = 6$, so $t = 1$ and the point is $(1, 2, 2)$. Check: $2 - 2 + 6 = 6$. Notice that the coefficient of $t$ after substitution, $4$, is exactly $\mathbf{d} \cdot \mathbf{n}$. That is not a coincidence, and it is the whole explanation of the three cases.

b) $\mathbf{d} \cdot \mathbf{n} = 2 - 2 + 0 = 0$. Substituting $(t, 2t, 0)$ gives $2t - 2t + 0 = 0 = 6$, an equation with no solution: the $t$ has disappeared because its coefficient is $\mathbf{d} \cdot \mathbf{n} = 0$, and what remains is false. The line is PARALLEL to $\Pi$ and does not meet it. Geometrically, $\mathbf{d}$ is orthogonal to the normal, so the line runs alongside the plane; the origin is at the wrong level, $0 \neq 6$.

c) $\mathbf{d} \cdot \mathbf{n} = 2 + 1 - 3 = 0$ again, so $L_3$ is parallel to $\Pi$. Substituting $(3 + t, -t, -t)$: $6 + 2t + t - 3t = 6$, that is $6 = 6$, true for EVERY $t$. The line lies inside the plane. Parts b) and c) have the same direction test and opposite conclusions, which is why the test alone never finishes the question: when $\mathbf{d} \cdot \mathbf{n} = 0$, check whether ONE point of the line satisfies the equation.

d) Parallel means $\mathbf{d} \cdot \mathbf{n} = 2a - 2 + 3 = 2a + 1 = 0$, so $a = -\frac{1}{2}$. The point $(1, 2, 0)$ gives $2 - 2 + 0 = 0 \neq 6$, so the line is parallel to $\Pi$ and disjoint from it. The trap on this part is the word parallel: $\mathbf{d} \cdot \mathbf{n} = 0$ means $\mathbf{d}$ is perpendicular to the NORMAL, hence parallel to the PLANE. A student who reads it as the line is perpendicular to the plane writes the exact opposite of the truth, and every later part inherits the error.

e) A line perpendicular to $\Pi$ has the direction of the NORMAL: $(1, 2, 2) + s(2, -1, 3)$. Here $\mathbf{d} \cdot \mathbf{n} = \|\mathbf{n}\|^2 = 14 \neq 0$, so it does cross $\Pi$, at $s = 0$, as it should. The two conditions to keep apart: a line is PERPENDICULAR to a plane when $\mathbf{d}$ is a multiple of $\mathbf{n}$, and PARALLEL to it when $\mathbf{d} \cdot \mathbf{n} = 0$.

a) $2(0) - (-5) + 2(0) = 5$, so $P$ is on $\Pi$. Then $\overrightarrow{PQ} = (3, 5, 4)$ and $\overrightarrow{PQ} \cdot \mathbf{n} = 6 - 5 + 8 = 9$, with $\|\mathbf{n}\| = \sqrt{4 + 1 + 4} = 3$. The projection of $\overrightarrow{PQ}$ on $\mathbf{n}$ is $\dfrac{9}{9}\mathbf{n} = (2, -1, 2)$, of length $3$, and this length IS the distance: on the figure, the normal direction is the only direction in which moving changes your distance to the plane. Note the contrast with exercise 3. For a line, the distance was the part LEFT after projecting on the direction; for a plane, it is the projection itself, on the normal. And $\|\overrightarrow{PQ}\| = \sqrt{50}$ is not the distance: $P$ was an arbitrary point.

b) Let $Q = (x_0, y_0, z_0)$ and let $P$ be any point of $ax + by + cz = d$, so $\mathbf{n} \cdot P = d$. Then $\overrightarrow{PQ} \cdot \mathbf{n} = \mathbf{n} \cdot Q - \mathbf{n} \cdot P = ax_0 + by_0 + cz_0 - d$, and the distance is the absolute value of the scalar component: $\dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$. The choice of $P$ has vanished from the result because $\mathbf{n} \cdot P = d$ for EVERY point of the plane; equivalently, two choices of $P$ differ by a vector lying in the plane, which is orthogonal to $\mathbf{n}$ and projects to zero. Here: $\dfrac{|6 - 0 + 8 - 5|}{3} = 3$. Forgetting the denominator, the most frequent error, gives $9$.

c) From $Q$, move back along the normal by the projection found in a): $F = Q - \dfrac{\mathbf{n} \cdot Q - d}{\|\mathbf{n}\|^2}\,\mathbf{n} = (3, 0, 4) - \dfrac{9}{9}(2, -1, 2) = (1, 1, 2)$. Check that $F$ is on $\Pi$: $2 - 1 + 4 = 5$. Check that $\overrightarrow{FQ} = (2, -1, 2)$ is parallel to $\mathbf{n}$: it is $\mathbf{n}$ itself, so $FQ$ meets the plane at a right angle, and $\|\overrightarrow{FQ}\| = 3$ as in a). Dividing by $\|\mathbf{n}\|$ here instead of $\|\mathbf{n}\|^2$ gives $(-3, 3, -2)$, which is not on $\Pi$ ($-6 - 3 - 4 = -13$), and the substitution catches it at once.

d) The mirror image is as far beyond $F$ as $Q$ is before it: $Q' = Q - 2(2, -1, 2) = (-1, 2, 0)$, and $F$ is the midpoint of $QQ'$, since $\frac{1}{2}(3 - 1, 0 + 2, 4 + 0) = (1, 1, 2)$. Distance check with the formula: $\dfrac{|-2 - 2 + 0 - 5|}{3} = \dfrac{9}{3} = 3$. The quantity inside the absolute value is $-9$ for $Q'$ and $+9$ for $Q$, which is the subject of e).

e) Drop the absolute value and look at the sign of $s(X) = 2x - y + 2z - 5$. $s(Q) = 9 > 0$, $s(0, 0, 0) = -5 < 0$: opposite signs, so the origin and $Q$ are on OPPOSITE sides and any segment joining them crosses $\Pi$. $s(R) = 2 - 1 + 2 - 5 = -2 < 0$, the sign of the origin: same side as the origin. The absolute value in the formula is exactly what throws this information away, so a question about sides is answered BEFORE taking it.

a) $\mathbf{n}_2 = (2, 4, -4) = 2\mathbf{n}_1$, so the planes are parallel. Divide the equation of $\Pi_2$ by $2$ to give it the SAME normal: $x + 2y - 2z = 9$. Same left side, different right side, so the planes are distinct. Now the distance is $\dfrac{|9 - 3|}{\|\mathbf{n}_1\|} = \dfrac{6}{3} = 2$. The classic wrong line is $\dfrac{|18 - 3|}{3} = 5$, comparing right-hand sides of equations written with different normals. A second route, which needs no rescaling, checks it: $(3, 0, 0)$ is on $\Pi_1$, and its distance to $\Pi_2$ is $\dfrac{|6 - 18|}{\sqrt{4 + 16 + 16}} = \dfrac{12}{6} = 2$.

b) A plane parallel to $\Pi_1$ is $x + 2y - 2z = c$, and its distance to $\Pi_1$ is $\dfrac{|c - 3|}{3}$. Setting this equal to $1$ gives $|c - 3| = 3$, so $c = 0$ or $c = 6$. There are TWO answers, one on each side, and giving only one loses half the part. A useful sanity check: $x + 2y - 2z = 0$ passes through the origin, and the distance from the origin to $\Pi_1$ is indeed $\frac{3}{3} = 1$.

c) $\mathbf{n}_1 = (1, 0, -1)$ and $\mathbf{n}_2 = (0, 1, -1)$: $\mathbf{n}_1 \cdot \mathbf{n}_2 = 1$ and both lengths are $\sqrt{2}$, so $\cos\theta = \dfrac{1}{2}$ and $\theta = \dfrac{\pi}{3}$. Why the normals: on the figure, turning plane $\Pi_1$ by $\theta$ onto plane $\Pi_2$ turns its normal by the same $\theta$, since each normal is glued at a right angle to its plane. Using direction vectors lying in the planes instead does not work: a plane contains infinitely many directions, and two of them chosen at random can make any angle at all.

d) $\mathbf{n}_1 = (1, 1, 0)$, $\mathbf{n}_2 = (-1, 0, 0)$: $\cos\varphi = \dfrac{-1}{\sqrt{2} \times 1} = -\dfrac{1}{\sqrt{2}}$, so the NORMALS make $\varphi = \dfrac{3\pi}{4}$. The planes make the acute angle $\theta = \pi - \varphi = \dfrac{\pi}{4}$. The same plane could have been written $x = -4$, with normal $(1, 0, 0)$, and the computation would then have given $\frac{\pi}{4}$ directly: the sign of a normal is arbitrary, so the angle between planes is defined by $\cos\theta = \dfrac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|}$, with the absolute value. Answering $135^{\circ}$ loses the part.

e) Perpendicular planes have perpendicular normals: $(k, 2, -1) \cdot (1, 2, -2) = k + 4 + 2 = k + 6 = 0$, so $k = -6$. Check: $(-6, 2, -1) \cdot (1, 2, -2) = -6 + 4 + 2 = 0$. Do not confuse with the condition for PARALLEL planes, which asks for proportional normals and would have no solution here, since the $z$-coefficients $-1$ and $-2$ force a ratio of $2$ that the $y$-coefficients $2$ and $2$ refuse.