Question ID: #696
For a triangle ABC, let $\vec{p}=\vec{BC}, \vec{q}=\vec{CA}$ and $\vec{r}=\vec{BA}$. If $|\vec{p}|=2\sqrt{3}, |\vec{q}|=2$ and $\cos \theta = \frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p}\times(\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to:
- (1) 340
- (2) 220
- (3) 410
- (4) 200
Solution:
From the triangle law of vector addition:
$$\vec{BC} + \vec{CA} = \vec{BA} \implies \vec{p} + \vec{q} = \vec{r}$$
The angle between vectors $\vec{p}$ and $\vec{q}$ is $\theta$.
Given $\cos \theta = \frac{1}{\sqrt{3}}$, we have $\sin^2 \theta = 1 – \cos^2 \theta = 1 – \frac{1}{3} = \frac{2}{3}$.
Applying the cosine rule to the triangle (using the interior angle $\pi – \theta$):
$$|\vec{r}|^2 = |\vec{p}|^2 + |\vec{q}|^2 – 2|\vec{p}||\vec{q}|\cos(\pi – \theta)$$
$$|\vec{r}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2|\vec{p}||\vec{q}|\cos \theta$$
Substitute the given values:
$$|\vec{r}|^2 = (2\sqrt{3})^2 + (2)^2 + 2(2\sqrt{3})(2)\left(\frac{1}{\sqrt{3}}\right)$$
$$|\vec{r}|^2 = 12 + 4 + 8 = 24$$
Now evaluate the required expression:
$$E = |\vec{p}\times(\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2}$$
Substitute $\vec{r} = \vec{p} + \vec{q}$:
$$E = |\vec{p}\times(\vec{q}-3(\vec{p}+\vec{q}))|^{2}+3(24)$$
$$E = |\vec{p}\times(\vec{q}-3\vec{p}-3\vec{q})|^{2}+72$$
$$E = |\vec{p}\times(-3\vec{p}-2\vec{q})|^{2}+72$$
Using properties of cross product ($\vec{p} \times \vec{p} = 0$):
$$E = |-2(\vec{p}\times\vec{q})|^{2}+72$$
$$E = 4|\vec{p}\times\vec{q}|^{2}+72$$
$$E = 4(|\vec{p}| |\vec{q}| \sin \theta)^2 + 72$$
$$E = 4|\vec{p}|^2 |\vec{q}|^2 \sin^2 \theta + 72$$
$$E = 4(12)(4)\left(\frac{2}{3}\right) + 72$$
$$E = 192\left(\frac{2}{3}\right) + 72$$
$$E = 64(2) + 72 = 128 + 72 = 200$$
Ans. (4)
Was this solution helpful?
YesNo