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Question: Let three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ be such that. $\vec{a}\times\vec{b}=3(\vec{a}\t...

Let three vectors a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} be such that. a⃗×b⃗=3(a⃗×c⃗),∣a⃗∣=1,∣b⃗∣=1,∣c⃗∣=13\vec{a}\times\vec{b}=3(\vec{a}\times\vec{c}), |\vec{a}|=1, |\vec{b}|=1, |\vec{c}|=\frac{1}{3}. If angle between b⃗\vec{b} and c⃗\vec{c} is π3\frac{\pi}{3}, then angle between a⃗\vec{a} and c⃗\vec{c} can be

A

cos⁡−1(35)\cos^{-1}(\frac{3}{5})

B

π−cos⁡−1(35)\pi - \cos^{-1}(\frac{3}{5})

C

5π6\frac{5\pi}{6}

D

π3\frac{\pi}{3}

Answer

π3\frac{\pi}{3}

Explanation

Solution

The given vector equation is a⃗×b⃗=3(a⃗×c⃗)\vec{a}\times\vec{b}=3(\vec{a}\times\vec{c}). Rearranging this, we get a⃗×b⃗−3a⃗×c⃗=0⃗\vec{a}\times\vec{b} - 3\vec{a}\times\vec{c} = \vec{0}, which can be written as a⃗×(b⃗−3c⃗)=0⃗\vec{a}\times(\vec{b}-3\vec{c}) = \vec{0}.

This implies that vector a⃗\vec{a} is parallel to the vector (b⃗−3c⃗)(\vec{b}-3\vec{c}). Let v⃗=b⃗−3c⃗\vec{v} = \vec{b}-3\vec{c}. Then a⃗∥v⃗\vec{a} \parallel \vec{v}.

We calculate the magnitude of v⃗\vec{v}: ∣v⃗∣2=∣b⃗−3c⃗∣2=(b⃗−3c⃗)⋅(b⃗−3c⃗)|\vec{v}|^2 = |\vec{b}-3\vec{c}|^2 = (\vec{b}-3\vec{c})\cdot(\vec{b}-3\vec{c}) ∣v⃗∣2=∣b⃗∣2−3b⃗⋅c⃗−3c⃗⋅b⃗+9∣c⃗∣2|\vec{v}|^2 = |\vec{b}|^2 - 3\vec{b}\cdot\vec{c} - 3\vec{c}\cdot\vec{b} + 9|\vec{c}|^2 ∣v⃗∣2=∣b⃗∣2−6(b⃗⋅c⃗)+9∣c⃗∣2|\vec{v}|^2 = |\vec{b}|^2 - 6(\vec{b}\cdot\vec{c}) + 9|\vec{c}|^2

We are given ∣b⃗∣=1|\vec{b}|=1, ∣c⃗∣=13|\vec{c}|=\frac{1}{3}, and the angle between b⃗\vec{b} and c⃗\vec{c} is π3\frac{\pi}{3}. So, b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡(π3)=(1)(13)(12)=16\vec{b}\cdot\vec{c} = |\vec{b}||\vec{c}|\cos(\frac{\pi}{3}) = (1)(\frac{1}{3})(\frac{1}{2}) = \frac{1}{6}.

Substitute these values into the expression for ∣v⃗∣2|\vec{v}|^2: ∣v⃗∣2=(1)2−6(16)+9(13)2|\vec{v}|^2 = (1)^2 - 6(\frac{1}{6}) + 9(\frac{1}{3})^2 ∣v⃗∣2=1−1+9(19)=1|\vec{v}|^2 = 1 - 1 + 9(\frac{1}{9}) = 1 So, ∣v⃗∣=∣b⃗−3c⃗∣=1|\vec{v}| = |\vec{b}-3\vec{c}| = 1.

Since a⃗∥(b⃗−3c⃗)\vec{a} \parallel (\vec{b}-3\vec{c}) and ∣a⃗∣=1|\vec{a}|=1 and ∣b⃗−3c⃗∣=1|\vec{b}-3\vec{c}|=1, there are two possibilities:

  1. a⃗=b⃗−3c⃗\vec{a} = \vec{b}-3\vec{c}
  2. a⃗=−(b⃗−3c⃗)=3c⃗−b⃗\vec{a} = -(\vec{b}-3\vec{c}) = 3\vec{c}-\vec{b}

Let θ\theta be the angle between a⃗\vec{a} and c⃗\vec{c}. We have a⃗⋅c⃗=∣a⃗∣∣c⃗∣cos⁡θ=(1)(13)cos⁡θ=13cos⁡θ\vec{a}\cdot\vec{c} = |\vec{a}||\vec{c}|\cos\theta = (1)(\frac{1}{3})\cos\theta = \frac{1}{3}\cos\theta.

Case 1: a⃗=b⃗−3c⃗\vec{a} = \vec{b}-3\vec{c} Take the dot product with c⃗\vec{c}: a⃗⋅c⃗=(b⃗−3c⃗)⋅c⃗=b⃗⋅c⃗−3∣c⃗∣2\vec{a}\cdot\vec{c} = (\vec{b}-3\vec{c})\cdot\vec{c} = \vec{b}\cdot\vec{c} - 3|\vec{c}|^2 13cos⁡θ=16−3(13)2=16−3(19)=16−13=−16\frac{1}{3}\cos\theta = \frac{1}{6} - 3(\frac{1}{3})^2 = \frac{1}{6} - 3(\frac{1}{9}) = \frac{1}{6} - \frac{1}{3} = -\frac{1}{6} cos⁡θ=−36=−12\cos\theta = -\frac{3}{6} = -\frac{1}{2} This gives θ=2π3\theta = \frac{2\pi}{3}. This angle is not present in the options.

Case 2: a⃗=3c⃗−b⃗\vec{a} = 3\vec{c}-\vec{b} Take the dot product with c⃗\vec{c}: a⃗⋅c⃗=(3c⃗−b⃗)⋅c⃗=3∣c⃗∣2−b⃗⋅c⃗\vec{a}\cdot\vec{c} = (3\vec{c}-\vec{b})\cdot\vec{c} = 3|\vec{c}|^2 - \vec{b}\cdot\vec{c} 13cos⁡θ=3(13)2−16=3(19)−16=13−16=16\frac{1}{3}\cos\theta = 3(\frac{1}{3})^2 - \frac{1}{6} = 3(\frac{1}{9}) - \frac{1}{6} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6} cos⁡θ=36=12\cos\theta = \frac{3}{6} = \frac{1}{2} This gives θ=π3\theta = \frac{\pi}{3}.

We check consistency with the initial condition ∣a⃗×b⃗∣=3∣a⃗×c⃗∣|\vec{a}\times\vec{b}|=3|\vec{a}\times\vec{c}|. ∣a⃗∣∣b⃗∣sin⁡θab=3∣a⃗∣∣c⃗∣sin⁡θac|\vec{a}||\vec{b}|\sin\theta_{ab} = 3|\vec{a}||\vec{c}|\sin\theta_{ac} 1⋅1⋅sin⁡θab=3⋅1⋅13⋅sin⁡θac1 \cdot 1 \cdot \sin\theta_{ab} = 3 \cdot 1 \cdot \frac{1}{3} \cdot \sin\theta_{ac} sin⁡θab=sin⁡θac\sin\theta_{ab} = \sin\theta_{ac}

In Case 2, θac=π3\theta_{ac} = \frac{\pi}{3}, so sin⁡θac=sin⁡(π3)=32\sin\theta_{ac} = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}. We need to check if sin⁡θab=32\sin\theta_{ab} = \frac{\sqrt{3}}{2}. From a⃗=3c⃗−b⃗\vec{a} = 3\vec{c}-\vec{b}, take the dot product with b⃗\vec{b}: a⃗⋅b⃗=(3c⃗−b⃗)⋅b⃗=3c⃗⋅b⃗−∣b⃗∣2\vec{a}\cdot\vec{b} = (3\vec{c}-\vec{b})\cdot\vec{b} = 3\vec{c}\cdot\vec{b} - |\vec{b}|^2 ∣a⃗∣∣b⃗∣cos⁡θab=3∣b⃗∣∣c⃗∣cos⁡(π3)−∣b⃗∣2|\vec{a}||\vec{b}|\cos\theta_{ab} = 3|\vec{b}||\vec{c}|\cos(\frac{\pi}{3}) - |\vec{b}|^2 (1)(1)cos⁡θab=3(1)(13)(12)−(1)2=12−1=−12(1)(1)\cos\theta_{ab} = 3(1)(\frac{1}{3})(\frac{1}{2}) - (1)^2 = \frac{1}{2} - 1 = -\frac{1}{2} cos⁡θab=−12\cos\theta_{ab} = -\frac{1}{2}. This implies θab=2π3\theta_{ab} = \frac{2\pi}{3}. Then sin⁡θab=sin⁡(2π3)=32\sin\theta_{ab} = \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}. This is consistent with sin⁡θac=32\sin\theta_{ac} = \frac{\sqrt{3}}{2}. Thus, θac=π3\theta_{ac} = \frac{\pi}{3} is a valid solution.