Vectors: Step-by-Step Worked Examples (4)

easy • 3 min read
Tags Vectors

Question

Find a unit vector perpendicular to both a⃗=2i^+j^−k^\vec{a} = 2\hat{i} + \hat{j} - \hat{k} and b⃗=i^−j^+2k^\vec{b} = \hat{i} - \hat{j} + 2\hat{k}. Then find the area of the parallelogram having a⃗\vec{a} and b⃗\vec{b} as adjacent sides. CBSE 2024 board pattern, also asked in JEE Main 2023.

Solution — Step by Step

The cross product a⃗×b⃗\vec{a} \times \vec{b} is perpendicular to both vectors. Using the determinant form:

a⃗×b⃗=∣i^j^k^21−11−12∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & 2 \end{vmatrix}

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

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

∣a⃗×b⃗∣=12+(−5)2+(−3)2=1+25+9=35|\vec{a} \times \vec{b}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}

n^=a⃗×b⃗∣a⃗×b⃗∣=135(i^−5j^−3k^)\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{1}{\sqrt{35}}(\hat{i} - 5\hat{j} - 3\hat{k})

The magnitude of the cross product equals the area of the parallelogram with the two vectors as adjacent sides.

Area=∣a⃗×b⃗∣=35≈5.92 sq. units\text{Area} = |\vec{a} \times \vec{b}| = \sqrt{35} \approx 5.92 \text{ sq. units}

Final answers: n^=135(i^−5j^−3k^)\hat{n} = \dfrac{1}{\sqrt{35}}(\hat{i} - 5\hat{j} - 3\hat{k}), Area =35= \sqrt{35} sq. units.

Why This Works

The cross product of two non-parallel vectors gives a third vector perpendicular to both. Its magnitude equals the area of the parallelogram they span — that’s not a coincidence, it’s the definition: ∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}||\vec{b}|\sin\theta is exactly the area formula for a parallelogram with sides ∣a⃗∣,∣b⃗∣|\vec{a}|, |\vec{b}| and included angle θ\theta.

The negative sign of the unit vector (−n^-\hat{n}) also works — it points the other way along the perpendicular axis. Either is acceptable in CBSE; JEE specifies “in the direction of a⃗×b⃗\vec{a} \times \vec{b}” if it wants a definite sign.

Alternative Method

Verify perpendicularity by computing dot products: n^⋅a⃗=(1)(2)+(−5)(1)+(−3)(−1)=2−5+3=0\hat{n} \cdot \vec{a} = (1)(2) + (-5)(1) + (-3)(-1) = 2 - 5 + 3 = 0. Similarly n^⋅b⃗=1+5−6=0\hat{n} \cdot \vec{b} = 1 + 5 - 6 = 0. Both zero confirms our perpendicular is correct.

The sign-flip mistake is universal. Computing i^(1⋅2−(−1)(−1))\hat{i}(1 \cdot 2 - (-1)(-1)) — many students drop the second negative and get 2−1=12 - 1 = 1 correctly, but the −j^-\hat{j} term often gets mishandled. Write out all signs explicitly.

For the area of a triangle with two sides a⃗\vec{a} and b⃗\vec{b}, halve the cross product: Area =12∣a⃗×b⃗∣= \frac{1}{2}|\vec{a} \times \vec{b}|. The parallelogram is twice the triangle.

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