Vector d Perpendicular to Both a and b, With c·d=15

NCERT Class 12 Mathematics — Vector Algebra, Miscellaneous Exercise, Question 12.

Problem

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 such that $\vec{c}\cdot\vec{d}=15$.

Key insight. “Perpendicular to both $\vec{a}$ and $\vec{b}$” is exactly what the cross product $\vec{a}\times\vec{b}$ gives — any vector perpendicular to both must be a scalar multiple of it. That collapses the problem from “find three unknown components” down to “find one unknown scalar,” which the condition $\vec{c}\cdot\vec{d}=15$ then pins down directly.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Since $\vec{d}$ must be perpendicular to both $\vec{a}$ and $\vec{b}$, write $\vec{d} = t(\vec{a}\times\vec{b})$ for an unknown scalar $t$. Compute the cross product, then use the condition $\vec{c}\cdot\vec{d}=15$ to solve for $t$.

Solution

Step 1 — Compute a × b

$$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7\end{vmatrix}$$

$$= \hat{i}\big(4\times7 – 2\times(-2)\big) – \hat{j}\big(1\times7-2\times3\big) + \hat{k}\big(1\times(-2)-4\times3\big)$$

$$= \hat{i}(28+4) – \hat{j}(7-6) + \hat{k}(-2-12) = 32\hat{i}-\hat{j}-14\hat{k}$$

Step 2 — Write d as a scalar multiple

Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, it must be parallel to $\vec{a}\times\vec{b}$:

$$\vec{d} = t(32\hat{i}-\hat{j}-14\hat{k})$$

Step 3 — Apply the condition c·d = 15

$$\vec{c}\cdot\vec{d} = t\big[(2)(32)+(-1)(-1)+(4)(-14)\big] = t(64+1-56) = 9t$$

Setting this equal to $15$:

$$9t = 15 \implies t = \frac{5}{3}$$

Step 4 — Write out d

$$\vec{d} = \frac{5}{3}(32\hat{i}-\hat{j}-14\hat{k}) = \frac{160}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{70}{3}\hat{k}$$

Answer

Common mistakes

• Trying to solve for the three components of $\vec{d}$ directly using three simultaneous equations. This works too (perpendicular to $\vec{a}$, perpendicular to $\vec{b}$, and the dot product condition give three linear equations), but recognising $\vec{d}\parallel\vec{a}\times\vec{b}$ immediately cuts the problem down to a single unknown.
• Sign errors in the cross product’s cofactor expansion. The middle term of the cross product carries a negative sign in front of it — easy to drop when computing quickly.

Practise next

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