Vectors: Common Mistakes and Fixes (1)

easy • 2 min read
Tags Vectors

Question

If a⃗=2i^+j^−3k^\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k} and b⃗=i^−2j^+k^\vec{b} = \hat{i} - 2\hat{j} + \hat{k}, find a unit vector perpendicular to both, and the area of the parallelogram with sides a⃗\vec{a} and b⃗\vec{b}.

Solution — Step by Step

a⃗×b⃗=∣i^j^k^21−31−21∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & -2 & 1 \end{vmatrix}

Expand: i^[(1)(1)−(−3)(−2)]−j^[(2)(1)−(−3)(1)]+k^[(2)(−2)−(1)(1)]\hat{i}[(1)(1) - (-3)(-2)] - \hat{j}[(2)(1) - (-3)(1)] + \hat{k}[(2)(-2) - (1)(1)].

=i^[1−6]−j^[2+3]+k^[−4−1]=−5i^−5j^−5k^= \hat{i}[1 - 6] - \hat{j}[2 + 3] + \hat{k}[-4 - 1] = -5\hat{i} - 5\hat{j} - 5\hat{k}.

 ∣a⃗×b⃗∣=25+25+25=53|\vec{a}\times\vec{b}| = \sqrt{25 + 25 + 25} = 5\sqrt{3} n^=a⃗×b⃗∣a⃗×b⃗∣=−5i^−5j^−5k^53=−13(i^+j^+k^)\hat{n} = \frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|} = \frac{-5\hat{i} - 5\hat{j} - 5\hat{k}}{5\sqrt{3}} = -\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})

The negative version 13(i^+j^+k^)\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) is also perpendicular — both are valid answers.

 Area=∣a⃗×b⃗∣=53 sq units\text{Area} = |\vec{a}\times\vec{b}| = 5\sqrt{3} \, \text{sq units}

Final answer: n^=±13(i^+j^+k^)\hat{n} = \pm\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}), area =53= 5\sqrt{3}.

Why This Works

The cross product a⃗×b⃗\vec{a} \times \vec{b} is perpendicular to both vectors by definition. Its magnitude equals the area of the parallelogram they span — this is the geometric meaning, not just an algebraic coincidence.

The right-hand rule fixes one direction; the opposite is also perpendicular, so the unit vector has a ±\pm ambiguity unless the problem specifies “in the direction of n^\hat{n} such that…”

Alternative Method

Sarrus rule for 3×3 determinants. Or expand along any row/column. The expansion along the first row is the most common.

The classic error: writing the determinant with rows interchanged or with a⃗\vec{a} in row 1 vs row 2. The order matters — a⃗×b⃗=−b⃗×a⃗\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}. Be consistent: question says ”a⃗\vec{a} and b⃗\vec{b}” — put a⃗\vec{a} in row 2 (just below the unit vectors) and b⃗\vec{b} in row 3.

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