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

A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is parallel to $\vec{a}\times\vec{b}$, so write $\vec{d} = \lambda(\vec{a}\times\vec{b})$ and fix $\lambda$ from $\vec{c}\cdot\vec{d}=15$.

Step 1 — Cross product.

Step 2 — Apply the dot-product condition. With $\vec{d} = \lambda(32\hat{i}-\hat{j}-14\hat{k})$ and $\vec{c} = 2\hat{i}-\hat{j}+4\hat{k}$,

$\vec{c}\cdot\vec{d} = \lambda(64+1-56) = 9\lambda = 15 \ \Rightarrow\ \lambda = \frac{5}{3}.$ …

The idea

Any vector perpendicular to both $\vec{a}$ and $\vec{b}$ must point along $\vec{a}\times\vec{b}$, because the cross product is itself perpendicular to both, and in 3-D the perpendiculars to two non-parallel vectors form a single line. So $\vec{d}$ can only be a scalar multiple of $\vec{a}\times\vec{b}$; the extra condition $\vec{c}\cdot\vec{d}=15$ pins down that scalar.

Step-by-step

1. Compute $\vec{a}\times\vec{b}$, with $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$, $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$:

• $\hat{i}$: $(4)(7)-(2)(-2) = 28+4 = 32$
• $\hat{j}$: $-\big[(1)(7)-(2)(3)\big] = -(7-6) = -1$
• $\hat{k}$: $(1)(-2)-(4)(3) = -2-12 = -14$

2. Write $\vec{d}$ as a multiple.

3. Use $\vec{c}\cdot\vec{d}=15$, with $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$: …

Method: A Vector Perpendicular to Two Given Vectors under a Scalar Condition

Use this when a vector must be perpendicular to two given vectors and also satisfy a dot-product value.

Steps

Step 1: Any vector perpendicular to both lies along their cross product

Since $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$, write

for an unknown scalar $\lambda$. Compute $\vec a\times\vec b$ via the determinant.

Step 2: Impose the scalar (dot-product) condition

Apply the given condition, e.g. $\vec c\cdot\vec d=15$: …

Common Mistakes

Mistake 1: Solving three scalar equations instead of using the cross product

Why it's wrong: setting $\vec d=(x,y,z)$ with $\vec d\cdot\vec a=0$, $\vec d\cdot\vec b=0$, $\vec c\cdot\vec d=15$ works but is slow and error-prone. Correct approach: recognise $\vec d\parallel\vec a\times\vec b$ and reduce to a single unknown $\lambda$.

Mistake 2: Assuming $\vec d$ is a unit vector or equals $\vec a\times\vec b$ itself

Why it's wrong: the condition $\vec c\cdot\vec d=15$ fixes the length, generally giving a non-unit multiple. Correct approach: keep the free scalar $\lambda$ and let the condition determine it. …