Q.If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}$, find a vector $\vec{c}$ such that $\vec{a}\times\vec{c}=\vec{b}$ and $\vec{a}\cdot\vec{c}=3$.

Key idea: write $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$, turn $\vec{a}\times\vec{c}=\vec{b}$ and $\vec{a}\cdot\vec{c}=3$ into equations, and solve.

Step 1 — Cross product. With $\vec{a} = \hat{i}+\hat{j}+\hat{k}$,

Matching components: $z - y = 0$, $x - z = 1$, $y - x = -1$ (the third repeats the first two).

Step 2 — Dot product. $x + y + z = 3$. …

We must find a vector $\vec{c}$ satisfying two conditions. A cross-product condition alone fixes only the part of $\vec{c}$ perpendicular to $\vec{a}$; the extra dot-product condition pins down the part along $\vec{a}$. Together they determine $\vec{c}$ uniquely.

1. Set up unknown components

Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. Here $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$.

2. Cross-product condition

Setting this equal to $\vec{b} = 0\hat{i} + \hat{j} - \hat{k}$ gives

Equation $(3)$ is just $(1)$ and $(2)$ combined, so it carries no new information.

3. Dot-product condition

4. Solve the system …

Method: Solving for an Unknown Vector from Cross- and Dot-Product Conditions

Use this whenever an unknown vector $\vec{c}$ must satisfy a cross-product equation $\vec{a}\times\vec{c}=\vec{b}$ together with a dot-product equation $\vec{a}\cdot\vec{c}=k$.

Steps

Step 1: Introduce components

Write $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$, turning the vector conditions into ordinary scalar equations in $x,y,z$.

Step 2: Convert the cross product into component equations

Expand $\vec{a}\times\vec{c}$ with the determinant and equate it component-by-component to $\vec{b}$. This gives three equations, but they are dependent — the cross product is always perpendicular to $\vec{a}$ — so only two are independent.

Step 3: Add the dot-product equation for the third constraint …

Common Mistakes

Mistake 1: Getting the cross-product order (and hence sign) wrong

Why it's wrong: $\vec{a}\times\vec{c}=-(\vec{c}\times\vec{a})$, so building the determinant with the rows swapped flips every component and gives $-\vec{b}$. Correct approach: keep $\vec{a}$ in the second row and $\vec{c}$ in the third to match $\vec{a}\times\vec{c}$.

Mistake 2: Treating all three cross-product component equations as independent

Why it's wrong: the cross product is always perpendicular to $\vec{a}$, so its three scalar equations are dependent — one is redundant. Correct approach: use two of them together with the dot-product equation. …