Vectors: Exam-Pattern Drill (6)

hard • 2 min read
Tags Vectors

Question

If a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} and b⃗=i^−j^+2k^\vec{b} = \hat{i} - \hat{j} + 2\hat{k}, find a unit vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}.

Solution — Step by Step

Any vector perpendicular to both a⃗\vec{a} and b⃗\vec{b} is a scalar multiple of a⃗×b⃗\vec{a} \times \vec{b}. So compute the cross product, then normalise.

 a⃗×b⃗=∣i^j^k^23−11−12∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix}

Expand: i^(3⋅2−(−1)(−1))−j^(2⋅2−(−1)(1))+k^(2⋅(−1)−3⋅1)\hat{i}(3 \cdot 2 - (-1)(-1)) - \hat{j}(2 \cdot 2 - (-1)(1)) + \hat{k}(2 \cdot (-1) - 3 \cdot 1).

=i^(6−1)−j^(4+1)+k^(−2−3)=5i^−5j^−5k^= \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) = 5\hat{i} - 5\hat{j} - 5\hat{k}.

 ∣a⃗×b⃗∣=25+25+25=75=53|\vec{a} \times \vec{b}| = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} n^=5i^−5j^−5k^53=13(i^−j^−k^)\hat{n} = \frac{5\hat{i} - 5\hat{j} - 5\hat{k}}{5\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} - \hat{k})

Final answer: n^=13(i^−j^−k^)\hat{n} = \tfrac{1}{\sqrt{3}}(\hat{i} - \hat{j} - \hat{k}) (and its negative is also valid).

Why This Works

The cross product is engineered to produce a vector perpendicular to both inputs. Its magnitude equals the area of the parallelogram spanned by a⃗\vec{a} and b⃗\vec{b}, which is why we divide by it to get a unit vector.

For CBSE Class 12 boards, this exact pattern appears almost every year in the 4-mark or 5-mark vector questions.

Alternative Method

Solve n⃗⋅a⃗=0\vec{n} \cdot \vec{a} = 0 and n⃗⋅b⃗=0\vec{n} \cdot \vec{b} = 0 as two equations in three unknowns, then normalise. Slower but useful if the cross product determinant looks ugly.

Watch for sign errors in the j^\hat{j} component — the alternating sign (+,−,+)(+, -, +) in determinant expansion catches half the class.

Common Mistake

Forgetting to normalise. The question asks for a “unit vector”, not just any perpendicular. Marks are deducted if the magnitude is not exactly 11.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Cross Product — Find a⃗ × b⃗ Using Determinant Method

medium

Dot Product of Two Vectors — Find Angle Between Them

easy

Find the area of parallelogram with adjacent sides a=2i+3j and b=i-4j

hard

If |a|=2, |b|=3 and a·b=4, find |a×b| and angle between vectors

easy

Find Magnitude of Vector 3î + 4ĵ

easy