Find Vector d Perpendicular to Both b and c, With d·a = 21

A standard vector algebra problem, not tied to a specific exam paper.

Problem

Given vectors $\vec{a} = 4\hat{i}+5\hat{j}-\hat{k}$, $\vec{b} = \hat{i}-4\hat{j}+5\hat{k}$, and $\vec{c} = 3\hat{i}+\hat{j}-\hat{k}$, find a vector $\vec{d}$ which is perpendicular to both $\vec{b}$ and $\vec{c}$, and $\vec{d} \cdot \vec{a} = 21$.

Key insight. “Perpendicular to two vectors” and “a specific dot product with a third” are just three linear conditions on the three unknown components of $\vec{d}$ — no need to reach for the cross product at all. Writing $\vec{d} = x\hat{i}+y\hat{j}+z\hat{k}$ and translating each condition directly into an equation turns this into ordinary simultaneous equations.

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Approach

Assume $\vec{d} = x\hat{i}+y\hat{j}+z\hat{k}$ in component form. Each condition in the problem — perpendicular to $\vec{b}$, perpendicular to $\vec{c}$, and a fixed dot product with $\vec{a}$ — becomes one linear equation in $x, y, z$, since a dot product of two vectors is just the sum of the products of matching components. Three conditions on three unknowns give a solvable system.

Solution

Step 1 — Translate $\vec{d}\cdot\vec{a}=21$ into an equation

$$(x)(4) + (y)(5) + (z)(-1) = 21 \implies 4x+5y-z=21 \quad \text{…(i)}$$

Step 2 — Translate $\vec{d}\perp\vec{b}$ (i.e. $\vec{d}\cdot\vec{b}=0$)

$$(x)(1)+(y)(-4)+(z)(5) = 0 \implies x-4y+5z=0 \quad \text{…(ii)}$$

Step 3 — Translate $\vec{d}\perp\vec{c}$ (i.e. $\vec{d}\cdot\vec{c}=0$)

$$(x)(3)+(y)(1)+(z)(-1) = 0 \implies 3x+y-z=0 \quad \text{…(iii)}$$

Step 4 — Solve the system

From (iii): $z = 3x+y$. Substituting into (ii):

$$x – 4y + 5(3x+y) = 0 \implies x-4y+15x+5y=0 \implies 16x+y=0 \implies y=-16x$$

Then $z = 3x + (-16x) = -13x$.

Substituting $y=-16x$ and $z=-13x$ into (i):

$$4x + 5(-16x) – (-13x) = 21 \implies 4x-80x+13x=21 \implies -63x=21 \implies x=-\frac{1}{3}$$

So:

$$y = -16\left(-\frac{1}{3}\right) = \frac{16}{3}, \qquad z = -13\left(-\frac{1}{3}\right) = \frac{13}{3}$$

Answer

Common mistakes

• Sign slips translating “perpendicular” into “dot product = 0”. It’s easy to write $\vec{d}\cdot\vec{b}$ with a component sign flipped, especially since $\vec{b}$ and $\vec{c}$ here both contain negative components.
• Substituting in the wrong order. Solving for $z$ in terms of $x$ first (from equation iii) works cleanly here because it avoids fractions early on; substituting a different equation first can make the algebra messier without being wrong.

Practise next

• Given $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$, $\vec{c}=\hat{i}-\hat{j}+2\hat{k}$, find $\vec{d}$ perpendicular to both $\vec{b}$ and $\vec{c}$ with $\vec{d}\cdot\vec{a}=10$, using the same system-of-equations method.