Vectors: PYQ Walkthrough (2)

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Tags Vectors

Question

(JEE Main style.) Let a⃗=2i^+j^−3k^\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k} and b⃗=i^−2j^+k^\vec{b} = \hat{i} - 2\hat{j} + \hat{k}. Find the projection of a⃗\vec{a} on b⃗\vec{b}, the angle between them, and the area of the parallelogram formed by a⃗\vec{a} and b⃗\vec{b}.

Solution — Step by Step

a⃗⋅b⃗=(2)(1)+(1)(−2)+(−3)(1)=2−2−3=−3\vec{a}\cdot\vec{b} = (2)(1) + (1)(-2) + (-3)(1) = 2 - 2 - 3 = -3

∣a⃗∣=4+1+9=14|\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14} and ∣b⃗∣=1+4+1=6|\vec{b}| = \sqrt{1 + 4 + 1} = \sqrt{6}.

projb⃗a⃗=a⃗⋅b⃗∣b⃗∣=−36=−62\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{-3}{\sqrt{6}} = -\frac{\sqrt{6}}{2}

The negative sign means a⃗\vec{a} has a component opposite to b⃗\vec{b}.

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−3146=−384=−3221\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{-3}{\sqrt{14}\sqrt{6}} = \dfrac{-3}{\sqrt{84}} = -\dfrac{3}{2\sqrt{21}}.

For the parallelogram area, compute a⃗×b⃗\vec{a}\times\vec{b}:

a⃗×b⃗=∣i^j^k^21−31−21∣=i^(1−6)−j^(2+3)+k^(−4−1)=−5i^−5j^−5k^\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-3\\1&-2&1\end{vmatrix} = \hat{i}(1-6) - \hat{j}(2+3) + \hat{k}(-4-1) = -5\hat{i} - 5\hat{j} - 5\hat{k}

∣a⃗×b⃗∣=53|\vec{a}\times\vec{b}| = 5\sqrt{3}. So area =53= 5\sqrt{3} square units.

Why This Works

Projection of a⃗\vec{a} on b⃗\vec{b} is a scalar — how much of a⃗\vec{a} “lies along” b⃗\vec{b}. The formula a⃗⋅b⃗/∣b⃗∣\vec{a}\cdot\vec{b}/|\vec{b}| comes from ∣a⃗∣cos⁡θ|\vec{a}|\cos\theta where θ\theta is the angle between them.

The cross product magnitude gives the area of the parallelogram because ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta — exactly the formula for parallelogram area in plane geometry.

When you see “projection,” think dot product divided by magnitude. When you see “area” or “perpendicular,” think cross product. Two reflexes that solve 60% of vector questions instantly.

Alternative Method

For the cross product, you can compute its components first (−5,−5,−5-5, -5, -5) and then take the magnitude. Or recognize that the cross product is along −i^−j^−k^-\hat{i}-\hat{j}-\hat{k}, so its magnitude is 535\sqrt{3}. Same answer.

Students confuse “projection of a⃗\vec{a} on b⃗\vec{b}” with “component of a⃗\vec{a} along b⃗\vec{b}.” They are the same scalar quantity but expressed differently. Don’t divide by ∣a⃗∣|\vec{a}| — divide by ∣b⃗∣|\vec{b}|.

Final answer: projection =−6/2= -\sqrt{6}/2, cos⁡θ=−3/(221)\cos\theta = -3/(2\sqrt{21}), area =53= 5\sqrt{3}.

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