Skip to content
Worked Examples · Example 23

Q.Find a unit vector perpendicular to each of the vectors a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b}, where a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}, b⃗=i^+2j^+3k^\vec{b}=\hat{i}+2\hat{j}+3\hat{k}.

Assam AhsecTextbookSubjective· 3mImportance★★★★★
Appeared in past exams:GUJCET 2025· Set 03· 1mexact
50% · 77/153 Questions
✓ Free question

The cross product (a⃗+b⃗)×(a⃗−b⃗)(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) gives a vector perpendicular to both, and normalizing it yields the unit vector 16(−i^+2j^−k^)\frac{1}{\sqrt{6}}(-\hat{i}+2\hat{j}-\hat{k}).

The key idea is simple: if you need a vector perpendicular to two given vectors, the cross product is your direct tool. Here, the two vectors are a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b}. Instead of computing these sums separately and then taking their cross product, we can use a neat property — the cross product simplifies to 2(b⃗×a⃗)2(\vec{b}\times\vec{a}), which saves work.

Let’s go step by step.

  1. Find a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b}. Given a⃗=i^+j^+k^\vec{a} = \hat{i}+\hat{j}+\hat{k} and b⃗=i^+2j^+3k^\vec{b} = \hat{i}+2\hat{j}+3\hat{k},

a⃗+b⃗=(1+1)i^+(1+2)j^+(1+3)k^=2i^+3j^+4k^\vec{a}+\vec{b} = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i}+3\hat{j}+4\hat{k}

a⃗−b⃗=(1−1)i^+(1−2)j^+(1−3)k^=0i^−1j^−2k^=−j^−2k^\vec{a}-\vec{b} = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} -1\hat{j} -2\hat{k} = -\hat{j}-2\hat{k}

  1. Compute the cross product (a⃗+b⃗)×(a⃗−b⃗)(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}). A vector perpendicular to both is given by their cross product. Let’s compute directly:

(a⃗+b⃗)×(a⃗−b⃗)=∣i^j^k^2340−1−2∣(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}

Expanding:

=i^∣34−1−2∣−j^∣240−2∣+k^∣230−1∣= \hat{i}\begin{vmatrix}3 & 4\\ -1 & -2\end{vmatrix} - \hat{j}\begin{vmatrix}2 & 4\\ 0 & -2\end{vmatrix} + \hat{k}\begin{vmatrix}2 & 3\\ 0 & -1\end{vmatrix}

=i^[3(−2)−4(−1)]−j^[2(−2)−4(0)]+k^[2(−1)−3(0)]= \hat{i}[3(-2) - 4(-1)] - \hat{j}[2(-2) - 4(0)] + \hat{k}[2(-1) - 3(0)]

=i^[−6+4]−j^[−4−0]+k^[−2−0]= \hat{i}[-6 + 4] - \hat{j}[-4 - 0] + \hat{k}[-2 - 0]

=i^(−2)−j^(−4)+k^(−2)=−2i^+4j^−2k^= \hat{i}(-2) - \hat{j}(-4) + \hat{k}(-2) = -2\hat{i} + 4\hat{j} - 2\hat{k}

So (a⃗+b⃗)×(a⃗−b⃗)=−2i^+4j^−2k^(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = -2\hat{i} + 4\hat{j} - 2\hat{k}.

Tip

You could also use the identity (a⃗+b⃗)×(a⃗−b⃗)=a⃗×a⃗−a⃗×b⃗+b⃗×a⃗−b⃗×b⃗=0−a⃗×b⃗+b⃗×a⃗−0=2(b⃗×a⃗)(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = \vec{a}\times\vec{a} - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - \vec{b}\times\vec{b} = 0 - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - 0 = 2(\vec{b}\times\vec{a}).

Computing b⃗×a⃗\vec{b}\times\vec{a} directly gives −i^+2j^−k^-\hat{i}+2\hat{j}-\hat{k}, and doubling it yields the same result. This shortcut avoids the determinant of the sum/difference vectors.

  1. Find the magnitude of this cross product.

∣−2i^+4j^−2k^∣=(−2)2+42+(−2)2=4+16+4=24=26| -2\hat{i} + 4\hat{j} - 2\hat{k} | = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}

  1. Normalize to get the unit vector. A unit vector perpendicular to both a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b} is:

−2i^+4j^−2k^26=16(−i^+2j^−k^)\frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{1}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k})

Watch out

The negative of this vector, 16(i^−2j^+k^)\frac{1}{\sqrt{6}}(\hat{i} - 2\hat{j} + \hat{k}), is also a unit vector perpendicular to both. Both are correct; the problem likely expects one of them. Always check if the question asks for "a" unit vector (any one) or "the" unit vector (often the one with a specific sign convention).

✓Final answer

The required unit vector is 16(−i^+2j^−k^)\boxed{\frac{1}{\sqrt{6}}(-\hat{i}+2\hat{j}-\hat{k})} (or its negative).

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question · 10.6.3Example 22 — Find |veca×vecb|, if veca=2hati+hatj+3hatk and vecb=3hati+5hatj-2hatk.→Next question · 10.6.3Example 24 — Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) a…