Vectors: Common Mistakes and Fixes (13)

easy • 2 min read
Tags Vectors

Question

If a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} and b⃗=i^−2j^+3k^\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}, find (a) a⃗⋅b⃗\vec{a} \cdot \vec{b}, (b) a⃗×b⃗\vec{a} \times \vec{b}, (c) the projection of a⃗\vec{a} on b⃗\vec{b}, (d) the angle between them.

Solution — Step by Step

a⃗⋅b⃗=(2)(1)+(3)(−2)+(−1)(3)=2−6−3=−7\vec{a} \cdot \vec{b} = (2)(1) + (3)(-2) + (-1)(3) = 2 - 6 - 3 = -7.

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

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

=i^(9−2)−j^(6+1)+k^(−4−3)=7i^−7j^−7k^= \hat{i}(9 - 2) - \hat{j}(6 + 1) + \hat{k}(-4 - 3) = 7\hat{i} - 7\hat{j} - 7\hat{k}.

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

Projection: a⃗⋅b⃗∣b⃗∣=−714=−142\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \dfrac{-7}{\sqrt{14}} = -\dfrac{\sqrt{14}}{2}.

(Negative — the angle between them is obtuse.)

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

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−714⋅14=−714=−12\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{-7}{\sqrt{14} \cdot \sqrt{14}} = \dfrac{-7}{14} = -\dfrac{1}{2}.

So θ=120°\theta = 120°.

Final answers: a⃗⋅b⃗=−7\vec{a} \cdot \vec{b} = \mathbf{-7}; a⃗×b⃗=7i^−7j^−7k^\vec{a} \times \vec{b} = \mathbf{7\hat{i} - 7\hat{j} - 7\hat{k}}; projection =−14/2= \mathbf{-\sqrt{14}/2}; θ=120°\theta = \mathbf{120°}.

Why This Works

The dot product encodes the angle: a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta. Negative dot product → obtuse angle. Zero → perpendicular. Positive → acute.

The cross product magnitude encodes the sine of the angle and gives a vector perpendicular to both. Verify: ∣a⃗×b⃗∣=49+49+49=73|\vec{a} \times \vec{b}| = \sqrt{49 + 49 + 49} = 7\sqrt{3}, and ∣a⃗∣∣b⃗∣sin⁡120°=14×3/2=73|\vec{a}||\vec{b}|\sin 120° = 14 \times \sqrt{3}/2 = 7\sqrt{3}. Check.

Alternative Method

For projection, use a⃗⋅b^\vec{a} \cdot \hat{b} where b^=b⃗/∣b⃗∣\hat{b} = \vec{b}/|\vec{b}|. Same answer, slightly cleaner notation.

Common Mistake

The most common mistake: students compute “projection of a⃗\vec{a} on b⃗\vec{b}” but write the formula for “projection of b⃗\vec{b} on a⃗\vec{a}”. The two are different unless ∣a⃗∣=∣b⃗∣|\vec{a}| = |\vec{b}|. The on-vector goes in the denominator. Also, watch the sign of the middle term in the determinant — that minus sign is responsible for many lost marks.

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