Vectors: Speed-Solving Techniques (8)

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Tags Vectors

Question

If 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 angle between a⃗\vec{a} and b⃗\vec{b}, and the projection of a⃗\vec{a} on 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,∣b⃗∣=1+4+1=6|\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}, \quad |\vec{b}| = \sqrt{1 + 4 + 1} = \sqrt{6} cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−3146=−384=−3221\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-3}{\sqrt{14}\sqrt{6}} = \frac{-3}{\sqrt{84}} = \frac{-3}{2\sqrt{21}}

So θ=cos⁡−1(−3221)≈109.1°\theta = \cos^{-1}\left(\dfrac{-3}{2\sqrt{21}}\right) \approx 109.1°. The negative cosine tells us the angle is obtuse.

 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}‘s component along b⃗\vec{b} points opposite to b⃗\vec{b} — consistent with the obtuse angle we just found.

Final Answer: θ=cos⁡−1(−3221)\theta = \cos^{-1}\left(\dfrac{-3}{2\sqrt{21}}\right), projection =−62= -\dfrac{\sqrt{6}}{2}.

Why This Works

The dot product packs both magnitude and angle information into one scalar. Once we have a⃗⋅b⃗\vec{a} \cdot \vec{b} and the two magnitudes, the angle drops out instantly. Projection is just dot product divided by the magnitude of whichever vector we’re projecting onto.

For speed: compute the dot product first. If you only need the angle, two magnitudes finish the job. If you need scalar projection, you only need one magnitude (the target vector). Choose the formula that uses the least extra arithmetic.

Alternative Method

Component-wise: project each component of a⃗\vec{a} onto the unit vector b^=b⃗/∣b⃗∣\hat{b} = \vec{b}/|\vec{b}| and sum. Algebraically identical, but more steps. The dot-product shortcut wins.

Confusing scalar projection with vector projection: projb⃗a⃗\text{proj}_{\vec{b}} \vec{a} as a vector is a⃗⋅b⃗∣b⃗∣2b⃗\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} (note the squared magnitude). As a scalar, it is a⃗⋅b⃗∣b⃗∣\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}. Read the question carefully — JEE often asks for the vector form.

Sign of a⃗⋅b⃗\vec{a} \cdot \vec{b} tells you the angle category instantly: positive means acute, zero means perpendicular, negative means obtuse. Use this for one-mark MCQs where you don’t need the exact angle.

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