A standard vector algebra problem, not tied to a specific exam paper.
Problem
If the vectors $\vec a=\hat i-\hat j+2\hat k$, $\vec b=2\hat i+4\hat j+\hat k$ and $\vec c=\lambda\hat i+\hat j+\mu\hat k$ are mutually orthogonal, then $(\lambda,\mu)=$?
Key insight. “Mutually orthogonal” for three vectors means every pair has a zero dot product — but that’s three conditions for only two unknowns ($\lambda$ and $\mu$, since $\vec a$ and $\vec b$ are already fully fixed). Checking $\vec a\cdot\vec b$ first confirms it’s already zero on its own, leaving exactly two genuinely useful equations for the two unknowns.
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Approach
First check that $\vec a\cdot\vec b=0$ already holds (a consistency check, since $\vec a$ and $\vec b$ have no free parameters). Then write $\vec a\cdot\vec c=0$ and $\vec b\cdot\vec c=0$ as two linear equations in $\lambda$ and $\mu$, and solve the resulting system.
Solution
Step 1 — Verify a · b = 0
$$\vec a\cdot\vec b = (1)(2)+(-1)(4)+(2)(1) = 2-4+2 = 0 \checkmark$$
Step 2 — Set up a · c = 0
$$\vec a\cdot\vec c = (1)(\lambda)+(-1)(1)+(2)(\mu) = \lambda-1+2\mu = 0$$
$$\lambda+2\mu=1 \quad \text{…(i)}$$
Step 3 — Set up b · c = 0
$$\vec b\cdot\vec c = (2)(\lambda)+(4)(1)+(1)(\mu) = 2\lambda+4+\mu = 0$$
$$2\lambda+\mu=-4 \quad \text{…(ii)}$$
Step 4 — Solve the system
From (i): $\lambda=1-2\mu$. Substituting into (ii):
$$2(1-2\mu)+\mu=-4 \implies 2-4\mu+\mu=-4 \implies -3\mu=-6 \implies \mu=2$$
$$\lambda = 1-2(2) = -3$$
Step 5 — State the result
$$(\lambda,\mu) = (-3,2)$$
Answer
$$(\lambda,\mu) = (-3,\,2)$$
Common mistakes
- Assuming $\vec a\cdot\vec b=0$ needs solving for something. Since $\vec a$ and $\vec b$ have no unknowns, this dot product is either automatically zero (a sanity check that the problem is consistent) or it isn’t — there’s nothing to solve for here, unlike the two equations involving $\vec c$.
- Mixing up which equation came from which dot product when substituting. Keeping equations (i) and (ii) clearly labelled by which pair of vectors produced them avoids accidentally substituting into the wrong one.
Practise next
- If $\vec p=2\hat i+\hat j-\hat k$, $\vec q=\hat i+\hat j+3\hat k$ and $\vec r=\lambda\hat i+\mu\hat j+\hat k$ are mutually orthogonal, find $(\lambda,\mu)$, using the same pairwise dot-product approach.
Show answer
$(\lambda,\mu)=(4,-7)$. Check the given pair first: $\vec p\cdot\vec q=2+1-3=0$, so they are orthogonal, as the question needs.
Then $\vec r\cdot\vec p=2\lambda+\mu-1=0$ and $\vec r\cdot\vec q=\lambda+\mu+3=0$.
Subtracting the second from the first gives $\lambda-4=0$, so $\lambda=4$ and $\mu=-7$. Checking $\vec p\cdot\vec q$ before anything else is not a formality — if it were non-zero, no $\lambda,\mu$ could make the three mutually orthogonal.

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