Given a = 4i + 5j − k, b = i − 4j + 5k and c = 3i + j − k, Find a Vector d Which Is Perpendicular to Both b and c and Satisfies 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 satisfies $\vec d\cdot\vec a=21$.

Key insight. “Perpendicular to $\vec b$” and “perpendicular to $\vec c$” are each a dot-product-equals-zero condition — together with the given $\vec d\cdot\vec a=21$, that’s three linear equations in the three unknown components of $\vec d$, solvable directly without ever computing a cross product.

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Approach

Let $\vec d=x\hat i+y\hat j+z\hat k$. Translate each of the three given conditions — $\vec d\cdot\vec a=21$, $\vec d\cdot\vec b=0$, $\vec d\cdot\vec c=0$ — into a linear equation in $x$, $y$, $z$. Three equations in three unknowns pin down $\vec d$ uniquely; solve by elimination.

Solution

Step 1 — Write the three conditions as linear equations

Let $\vec d=x\hat i+y\hat j+z\hat k$.

$$\vec d\cdot\vec a=21: \quad 4x+5y-z=21 \quad \text{…(1)}$$

$$\vec d\cdot\vec b=0: \quad x-4y+5z=0 \quad \text{…(2)}$$

$$\vec d\cdot\vec c=0: \quad 3x+y-z=0 \quad \text{…(3)}$$

Step 2 — Express z from equation (3)

$$z = 3x+y \quad \text{…(3′)}$$

Step 3 — Substitute into equation (2)

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

Step 4 — Express z in terms of x

Substituting $y=-16x$ into (3′):

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

Step 5 — Substitute into equation (1) to solve for x

$$4x+5(-16x)-(-13x) = 21 \implies 4x-80x+13x=21 \implies -63x=21 \implies x=-\frac13$$

Step 6 — Back-substitute for y and z

$$y = -16\left(-\frac13\right) = \frac{16}3, \qquad z = -13\left(-\frac13\right) = \frac{13}3$$

Step 7 — Verify

$$\vec d\cdot\vec a = 4\left(-\frac13\right)+5\left(\frac{16}3\right)-\frac{13}3 = \frac{-4+80-13}3=\frac{63}3=21 \checkmark$$

$$\vec d\cdot\vec b = -\frac13-4\left(\frac{16}3\right)+5\left(\frac{13}3\right) = \frac{-1-64+65}3=0 \checkmark$$

$$\vec d\cdot\vec c = 3\left(-\frac13\right)+\frac{16}3-\frac{13}3 = \frac{-3+16-13}3=0 \checkmark$$

Answer

Common mistakes

• Sign errors when substituting z=3x+y into equation (2). With three negative coefficients floating around ($-4y$, $+5z$), it’s easy to drop a sign while expanding $5(3x+y)$ — always re-expand term by term rather than combining mentally.
• Skipping the verification step. Because this problem has no options to check against, a single arithmetic slip anywhere in the elimination can go unnoticed — substituting the final $\vec d$ back into all three original conditions, as in Step 7, is the only way to catch it.
• Forgetting that “perpendicular to both b and c” gives two separate equations, not one. It’s tempting to try to combine the two perpendicularity conditions into a single cross-product shortcut, but setting up $\vec d\cdot\vec b=0$ and $\vec d\cdot\vec c=0$ as two ordinary linear equations is simpler and less error-prone here.

Practise next

• Using the same vectors $\vec a$, $\vec b$, $\vec c$, find the vector $\vec e$ perpendicular to both $\vec b$ and $\vec c$ with $\vec e\cdot\vec a=42$.