Solved Examples

How to Use This Section

Twenty-five fully worked problems covering the whole chapter, arranged in five graded batches: unit vectors and components, then resultants and the section formula, then the dot product, then the cross product, and finally mixed miscellaneous-level problems. Attempt each on paper before reading the solution — the worked version is for checking, not for first contact.

A classification habit worth building from the very first example: before computing, name the tool. Lengths and angles → dot product. Perpendiculars and areas → cross product. Points dividing segments → section formula. "Parallel to" or "collinear" → proportional components. Most errors in this chapter are not calculation slips — they are wrong-tool choices made in the first line.

Key Point: marks are lost at setup, not at arithmetic. Writing $\overrightarrow{AB} = \vec{b} - \vec{a}$ (terminal minus initial) correctly, choosing the dot for an angle, and remembering the $\dfrac{1}{2}$ for a triangle's area account for nearly every mark in this chapter.

Batch 1 — Unit Vectors and Components (Easy)

Example 1: All unit vectors in the XY-plane

Write all the unit vectors in the XY-plane.

Solution:

1. Parametrise by the angle: a unit vector $\vec{r}$ in the XY-plane making angle $\theta$ with the positive $x$-axis has components $x = \cos\theta$, $y = \sin\theta$.
2. Check the magnitude: $|\vec{r}| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$. ✓

Answer: $\vec{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}$, $0 \leq \theta < 2\pi$ — as $\theta$ sweeps a full turn, the tip traces the unit circle and every planar direction appears exactly once.

Example 2: One specific direction

Write a unit vector in the XY-plane making an angle of $30^\circ$ with the positive $x$-axis.

Solution:

1. Substitute into Example 1's formula: $\vec{r} = \cos 30^\circ\,\hat{i} + \sin 30^\circ\,\hat{j}$.

Answer: $\vec{r} = \dfrac{\sqrt{3}}{2}\hat{i} + \dfrac{1}{2}\hat{j}$.

Example 3: Scaling to unit length

Find the value of $x$ for which $x\left(\hat{i} + \hat{j} + \hat{k}\right)$ is a unit vector.

Solution:

1. Magnitude condition: $\left|x\left(\hat{i}+\hat{j}+\hat{k}\right)\right| = |x|\sqrt{3} = 1$.
2. Solve: $|x| = \dfrac{1}{\sqrt{3}}$.

Answer: $x = \pm\dfrac{1}{\sqrt{3}}$ — both signs work; the two answers point in opposite directions along the same line.

Example 4: The joining vector, in general

Find the scalar components and the magnitude of the vector joining $P(x_1, y_1, z_1)$ to $Q(x_2, y_2, z_2)$.

Solution:

1. Terminal minus initial: $\overrightarrow{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.
2. Magnitude: $\left|\overrightarrow{PQ}\right| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Answer: components $(x_2 - x_1),\ (y_2 - y_1),\ (z_2 - z_1)$; magnitude as above — the distance formula is just the length of a joining vector.

Example 5: A two-leg walk

A girl walks $4$ km towards the west, then $3$ km in a direction $30^\circ$ east of north, and stops. Find her displacement from the starting point.

Solution:

1. Set axes: east $= \hat{i}$, north $= \hat{j}$. First leg: $-4\hat{i}$.
2. Second leg: "$30^\circ$ east of north" tilts from north towards east: $3\left(\sin 30^\circ\,\hat{i} + \cos 30^\circ\,\hat{j}\right) = \dfrac{3}{2}\hat{i} + \dfrac{3\sqrt{3}}{2}\hat{j}$.
3. Add: $\left(-4 + \dfrac{3}{2}\right)\hat{i} + \dfrac{3\sqrt{3}}{2}\hat{j} = -\dfrac{5}{2}\hat{i} + \dfrac{3\sqrt{3}}{2}\hat{j}$.
4. Magnitude: $\sqrt{\dfrac{25}{4} + \dfrac{27}{4}} = \sqrt{13}$ km.

Answer: $-\dfrac{5}{2}\hat{i} + \dfrac{3\sqrt{3}}{2}\hat{j}$, of magnitude $\sqrt{13}$ km — the angle in "east of north" measures from north, which is why sine goes with $\hat{i}$ here.

Batch 2 — Resultants, Collinearity, and the Section Formula (Easy-Medium)

Example 6: A vector of magnitude 5 along a resultant

Find a vector of magnitude $5$ units parallel to the resultant of $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$.

Solution:

1. Resultant: $\vec{a} + \vec{b} = 3\hat{i} + \hat{j}$, magnitude $\sqrt{10}$.
2. Normalise and scale: $\pm 5 \cdot \dfrac{3\hat{i} + \hat{j}}{\sqrt{10}} = \pm\dfrac{\sqrt{10}}{2}\left(3\hat{i} + \hat{j}\right)$.

Answer: $\pm\left(\dfrac{3\sqrt{10}}{2}\hat{i} + \dfrac{\sqrt{10}}{2}\hat{j}\right)$ — "parallel to" admits both signs.

Example 7: A unit vector parallel to a combination

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$, find a unit vector parallel to $2\vec{a} - \vec{b} + 3\vec{c}$.

Solution:

1. Combine componentwise: $2\vec{a} - \vec{b} + 3\vec{c} = (2 - 2 + 3)\hat{i} + (2 + 1 - 6)\hat{j} + (2 - 3 + 3)\hat{k} = 3\hat{i} - 3\hat{j} + 2\hat{k}$.
2. Magnitude: $\sqrt{9 + 9 + 4} = \sqrt{22}$.

Answer: $\pm\dfrac{1}{\sqrt{22}}\left(3\hat{i} - 3\hat{j} + 2\hat{k}\right)$.

Example 8: Collinear points and the dividing ratio

Show that $A(1, -2, -8)$, $B(5, 0, -2)$ and $C(11, 3, 7)$ are collinear, and find the ratio in which $B$ divides $AC$.

Solution:

1. Joining vectors: $\overrightarrow{AB} = 4\hat{i} + 2\hat{j} + 6\hat{k}$, $\overrightarrow{BC} = 6\hat{i} + 3\hat{j} + 9\hat{k}$.
2. Proportionality: $\overrightarrow{BC} = \dfrac{3}{2}\overrightarrow{AB}$ — parallel vectors sharing the point $B$: collinear. ✓
3. Ratio: $AB : BC = 2 : 3$, so $B$ divides $AC$ internally in the ratio $2 : 3$.
4. Check with the section formula: $\dfrac{2\vec{c} + 3\vec{a}}{5} = \dfrac{(22 + 3, \ 6 - 6, \ 14 - 24)}{5} = (5, 0, -2) = \vec{b}$. ✓

Answer: collinear, with $B$ dividing $AC$ in the ratio $2 : 3$ internally.

Example 9: External division, and a midpoint that appears for free

$P$ and $Q$ have position vectors $2\vec{a} + \vec{b}$ and $\vec{a} - 3\vec{b}$. Find the position vector of $R$ dividing $PQ$ externally in the ratio $1 : 2$, and show that $P$ is the midpoint of $RQ$.

Solution:

1. External formula with $m : n = 1 : 2$: $\vec{r} = \dfrac{1\cdot\left(\vec{a} - 3\vec{b}\right) - 2\left(2\vec{a} + \vec{b}\right)}{1 - 2} = \dfrac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}$.
2. Midpoint of $RQ$: $\dfrac{\left(3\vec{a} + 5\vec{b}\right) + \left(\vec{a} - 3\vec{b}\right)}{2} = \dfrac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b}$ — precisely $P$. ✓

Answer: $\vec{r} = 3\vec{a} + 5\vec{b}$, and $P$ is the midpoint of $RQ$ — external division in $1:2$ always places the original first point midway between the new point and the second.

Example 10: An angle that turns out straight

The points $A, B, C, D$ have position vectors $\hat{i} + \hat{j} + \hat{k}$, $2\hat{i} + 5\hat{j}$, $3\hat{i} + 2\hat{j} - 3\hat{k}$ and $\hat{i} - 6\hat{j} - \hat{k}$. Find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$, and deduce that they are collinear.

Solution:

1. Joining vectors: $\overrightarrow{AB} = \hat{i} + 4\hat{j} - \hat{k}$ (magnitude $3\sqrt{2}$), $\overrightarrow{CD} = -2\hat{i} - 8\hat{j} + 2\hat{k}$ (magnitude $6\sqrt{2}$).
2. Cosine: $\cos\theta = \dfrac{-2 - 32 - 2}{\left(3\sqrt{2}\right)\left(6\sqrt{2}\right)} = \dfrac{-36}{36} = -1$.
3. Read the angle: $\theta = \pi$.

Answer: $\theta = \pi$ — the vectors are anti-parallel, hence collinear. Faster still: $\overrightarrow{CD} = -2\overrightarrow{AB}$ by inspection.

Batch 3 — Dot Product (Medium)

Example 11: Perpendicular to the sum of the others

Let $\vec{a}, \vec{b}, \vec{c}$ satisfy $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{c}| = 5$, each perpendicular to the sum of the other two. Find $\left|\vec{a} + \vec{b} + \vec{c}\right|$.

Solution:

1. Write the given conditions as dots: $\vec{a}\cdot\left(\vec{b}+\vec{c}\right) = 0$, $\vec{b}\cdot\left(\vec{c}+\vec{a}\right) = 0$, $\vec{c}\cdot\left(\vec{a}+\vec{b}\right) = 0$.
2. Square the sum: $\left|\vec{a}+\vec{b}+\vec{c}\right|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2\left(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\right)$.
3. The cross terms are the given conditions in disguise: adding the three conditions gives $2\left(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\right) = 0$.
4. Finish: $\left|\vec{a}+\vec{b}+\vec{c}\right|^2 = 9 + 16 + 25 = 50$.

Answer: $\left|\vec{a}+\vec{b}+\vec{c}\right| = 5\sqrt{2}$.

Example 12: Evaluating a sum of dots

Three vectors with $|\vec{a}| = 2$, $|\vec{b}| = 3$, $|\vec{c}| = 4$ satisfy $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Evaluate $\mu = \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}$.

Solution:

1. Square the zero: $0 = \left|\vec{a}+\vec{b}+\vec{c}\right|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2\mu$.
2. Substitute: $0 = 4 + 9 + 16 + 2\mu = 29 + 2\mu$.

Answer: $\mu = -\dfrac{29}{2}$ — the "square the sum" identity again, run backwards.

Example 13: A unit-vector condition fixes a parameter

The scalar product of $\hat{i} + \hat{j} + \hat{k}$ with the unit vector along the sum of $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ equals $1$. Find $\lambda$.

Solution:

1. The sum: $\vec{s} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}$, with $|\vec{s}| = \sqrt{(2+\lambda)^2 + 40}$.
2. The condition: $\dfrac{\left(\hat{i}+\hat{j}+\hat{k}\right)\cdot\vec{s}}{|\vec{s}|} = \dfrac{\lambda + 6}{\sqrt{(2+\lambda)^2 + 40}} = 1$.
3. Square and solve: $(\lambda + 6)^2 = (\lambda + 2)^2 + 40 \Rightarrow 12\lambda + 36 = 4\lambda + 44 \Rightarrow 8\lambda = 8$.

Answer: $\lambda = 1$ (and it satisfies the unsquared equation, since $\lambda + 6 = 7 > 0$).

Example 14: Equally inclined to three perpendicular directions

If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitude, show that $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to each of them.

Solution:

1. Common data: let $|\vec{a}| = |\vec{b}| = |\vec{c}| = m$; all pairwise dots vanish.
2. Length of the sum: $\left|\vec{a}+\vec{b}+\vec{c}\right|^2 = 3m^2$, so $\left|\vec{a}+\vec{b}+\vec{c}\right| = m\sqrt{3}$.
3. Angle with $\vec{a}$: $\cos\theta_a = \dfrac{\left(\vec{a}+\vec{b}+\vec{c}\right)\cdot\vec{a}}{m\sqrt{3}\cdot m} = \dfrac{m^2}{\sqrt{3}m^2} = \dfrac{1}{\sqrt{3}}$ — and identically for $\vec{b}$ and $\vec{c}$.

Answer: each angle is $\cos^{-1}\dfrac{1}{\sqrt{3}}$ — the body-diagonal of a cube makes equal angles with its three edges.

Example 15: Equal diagonals detect a right angle

Prove that $\left|\vec{a} + \vec{b}\right| = \left|\vec{a} - \vec{b}\right|$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular (both nonzero).

Solution:

1. Square both sides: $|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2$.
2. Cancel: $4\,\vec{a}\cdot\vec{b} = 0 \iff \vec{a}\cdot\vec{b} = 0 \iff \vec{a} \perp \vec{b}$.

Answer: proved — geometrically, a parallelogram has equal diagonals exactly when it is a rectangle.

Batch 4 — Cross Product (Medium)

Example 16: Diagonal and area of a parallelogram

The adjacent sides of a parallelogram are $\vec{a} = 2\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} - 3\hat{k}$. Find a unit vector parallel to its diagonal, and its area.

Solution:

1. Diagonal (through the common vertex): $\vec{a} + \vec{b} = 3\hat{i} - 6\hat{j} + 2\hat{k}$, magnitude $\sqrt{9 + 36 + 4} = 7$; unit vector $\dfrac{1}{7}\left(3\hat{i} - 6\hat{j} + 2\hat{k}\right)$.
2. Area: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = (12 + 10)\hat{i} - (-6 - 5)\hat{j} + (-4 + 4)\hat{k} = 22\hat{i} + 11\hat{j}$.
3. Magnitude: $\sqrt{484 + 121} = 11\sqrt{5}$.

Answer: unit diagonal $\dfrac{1}{7}\left(3\hat{i} - 6\hat{j} + 2\hat{k}\right)$; area $11\sqrt{5}$ square units.

Example 17: Perpendicular to two vectors with a prescribed dot

Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$, $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$. Find a vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{b}$ with $\vec{c}\cdot\vec{d} = 15$.

Solution:

1. Perpendicular to both means parallel to the cross: $\vec{d} = \lambda\left(\vec{a} \times \vec{b}\right)$, and $\vec{a} \times \vec{b} = (28 + 4)\hat{i} - (7 - 6)\hat{j} + (-2 - 12)\hat{k} = 32\hat{i} - \hat{j} - 14\hat{k}$.
2. Apply the dot condition: $\vec{c}\cdot\left(\vec{a}\times\vec{b}\right) = 64 + 1 - 56 = 9$, so $9\lambda = 15$ and $\lambda = \dfrac{5}{3}$.

Answer: $\vec{d} = \dfrac{5}{3}\left(32\hat{i} - \hat{j} - 14\hat{k}\right) = \dfrac{1}{3}\left(160\hat{i} - 5\hat{j} - 70\hat{k}\right)$.

Example 18: Splitting a vector along and across a direction

With $\vec{\alpha} = 3\hat{i} - \hat{j}$ and $\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}$, express $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$ where $\vec{\beta}_1 \parallel \vec{\alpha}$ and $\vec{\beta}_2 \perp \vec{\alpha}$.

Solution:

1. Set up: $\vec{\beta}_1 = \lambda\vec{\alpha} = 3\lambda\hat{i} - \lambda\hat{j}$, so $\vec{\beta}_2 = \vec{\beta} - \lambda\vec{\alpha} = (2 - 3\lambda)\hat{i} + (1 + \lambda)\hat{j} - 3\hat{k}$.
2. Perpendicularity: $\vec{\beta}_2\cdot\vec{\alpha} = 3(2 - 3\lambda) - (1 + \lambda) = 5 - 10\lambda = 0$, so $\lambda = \dfrac{1}{2}$.
3. Read off both parts: $\vec{\beta}_1 = \dfrac{3}{2}\hat{i} - \dfrac{1}{2}\hat{j}$ and $\vec{\beta}_2 = \dfrac{1}{2}\hat{i} + \dfrac{3}{2}\hat{j} - 3\hat{k}$.

Answer: $\vec{\beta} = \left(\dfrac{3}{2}\hat{i} - \dfrac{1}{2}\hat{j}\right) + \left(\dfrac{1}{2}\hat{i} + \dfrac{3}{2}\hat{j} - 3\hat{k}\right)$ — resolution along and perpendicular to a direction, the decomposition JEE loves.

Example 19: The same angle by two roads

For $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$, find the angle between them using the dot product, and confirm it with the cross product.

Solution:

1. Dot route: $\vec{a}\cdot\vec{b} = 1$, $|\vec{a}||\vec{b}| = 2$, so $\cos\theta = \dfrac{1}{2}$ and $\theta = 60^\circ$.
2. Cross route: $\vec{a} \times \vec{b} = \hat{i} - \hat{j} + \hat{k}$, magnitude $\sqrt{3}$, so $\sin\theta = \dfrac{\sqrt{3}}{2}$ — again $\theta = 60^\circ$. ✓

Answer: $60^\circ$ by both routes. When only $\sin\theta$ is computed, remember it cannot distinguish $\theta$ from $\pi - \theta$ — the dot product's sign settles which; here the positive dot confirms the acute choice.

Example 20: An identity worth knowing

Show that $\left(\vec{a} - \vec{b}\right) \times \left(\vec{a} + \vec{b}\right) = 2\left(\vec{a} \times \vec{b}\right)$.

Solution:

1. Expand by distributivity: $\vec{a}\times\vec{a} + \vec{a}\times\vec{b} - \vec{b}\times\vec{a} - \vec{b}\times\vec{b}$.
2. Kill the self-crosses and flip the reversed one: $= \vec{0} + \vec{a}\times\vec{b} + \vec{a}\times\vec{b} - \vec{0}$.

Answer: $2\left(\vec{a}\times\vec{b}\right)$ — geometrically: the parallelogram on the two diagonals has twice the area of the one on the sides.

Batch 5 — Mixed Problems (Medium-Hard)

Example 21: Length of a combination

If $|\vec{a}| = 2$, $|\vec{b}| = 5$ and the angle between them is $60^\circ$, find $\left|2\vec{a} + \vec{b}\right|$.

Solution:

1. Dot first: $\vec{a}\cdot\vec{b} = 2 \times 5 \times \cos 60^\circ = 5$.
2. Expand: $\left|2\vec{a}+\vec{b}\right|^2 = 4|\vec{a}|^2 + 4\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 16 + 20 + 25 = 61$.

Answer: $\left|2\vec{a}+\vec{b}\right| = \sqrt{61}$.

Example 22: The fourth vertex of a parallelogram

Three vertices of parallelogram $ABCD$ (in order) are $A(1, 1, 1)$, $B(2, 3, 4)$, $C(4, 5, 6)$. Find $D$.

Solution:

1. Opposite sides equal: $\overrightarrow{AD} = \overrightarrow{BC}$, so $\vec{d} - \vec{a} = \vec{c} - \vec{b}$.
2. Solve: $\vec{d} = \vec{a} + \vec{c} - \vec{b} = (1 + 4 - 2, \ 1 + 5 - 3, \ 1 + 6 - 4) = (3, 3, 3)$.

Answer: $D(3, 3, 3)$. Equivalent check: the diagonals bisect each other — the midpoint of $AC$ is $\left(\frac{5}{2}, 3, \frac{7}{2}\right)$, which is also the midpoint of $BD$. ✓

Example 23: One length forces another

Unit vectors $\hat{a}$ and $\hat{b}$ satisfy $\left|\hat{a} + \hat{b}\right| = 1$. Find $\left|\hat{a} - \hat{b}\right|$.

Solution:

1. Square the given: $1 = 1 + 1 + 2\hat{a}\cdot\hat{b}$, so $\hat{a}\cdot\hat{b} = -\dfrac{1}{2}$ (the angle is $120^\circ$).
2. Square the target: $\left|\hat{a}-\hat{b}\right|^2 = 2 - 2\hat{a}\cdot\hat{b} = 2 + 1 = 3$.

Answer: $\left|\hat{a} - \hat{b}\right| = \sqrt{3}$ — the two diagonal lengths of a unit rhombus always satisfy $d_1^2 + d_2^2 = 4$.

Example 24: Projection, and the perpendicular remainder

For $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$, find the projection of $\vec{a}$ on $\vec{b}$, the vector component of $\vec{a}$ along $\vec{b}$, and the component perpendicular to $\vec{b}$.

Solution:

1. Projection (a number): $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \dfrac{2 + 2 + 2}{3} = 2$.
2. Component along $\vec{b}$ (a vector): $\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b} = \dfrac{6}{9}\vec{b} = \dfrac{2}{3}\left(2\hat{i} + \hat{j} + 2\hat{k}\right)$.
3. Perpendicular remainder: $\vec{a} - \dfrac{2}{3}\vec{b} = -\dfrac{1}{3}\hat{i} + \dfrac{4}{3}\hat{j} - \dfrac{1}{3}\hat{k}$; check $\left(-\dfrac{1}{3}, \dfrac{4}{3}, -\dfrac{1}{3}\right)\cdot(2, 1, 2) = 0$. ✓

Answer: projection $2$; along-component $\dfrac{2}{3}\vec{b}$; perpendicular component $\dfrac{1}{3}\left(-\hat{i} + 4\hat{j} - \hat{k}\right)$.

Example 25: Rhombus diagonals, in two lines

Prove by vectors that the diagonals of a rhombus are perpendicular.

Solution:

1. Set up: let the sides from one vertex be $\vec{a}$ and $\vec{b}$ with $|\vec{a}| = |\vec{b}|$ (that is what makes it a rhombus). The diagonals are $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$.
2. Dot the diagonals: $\left(\vec{a}+\vec{b}\right)\cdot\left(\vec{a}-\vec{b}\right) = |\vec{a}|^2 - |\vec{b}|^2 = 0$.

Answer: the dot product of the diagonals vanishes, so they are perpendicular — a two-line proof of a classical theorem, and the template for most vector-geometry proofs: encode the figure in $\vec{a}, \vec{b}$, then compute one dot or cross.