If p=(a+1) i+a j+a k, q=a i+(a+1) j+a k, r=a i+a j+(a+1) k (a R) are coplanar, and 3( p q)…

Question

If $\vec p=(a+1)\hat i+a\hat j+a\hat k$, $\vec q=a\hat i+(a+1)\hat j+a\hat k$, $\vec r=a\hat i+a\hat j+(a+1)\hat k$ ($a\in\mathbb R$) are coplanar, and $3(\vec p\cdot\vec q)^2-\lambda|\vec r	imes\vec q|^2=0$, then $\lambda$ is \_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

Coplanarity: $\begin{vmatrix}a+1&a&a\a&a+1&a\a&a&a+1\end{vmatrix}=0 \implies 3a+1=0 \implies a=-\dfrac13$

$\vec p=\dfrac13(2,-1,-1)$, $\vec q=\dfrac13(-1,2,-1)$, $\vec r=\dfrac13(-1,-1,2)$.

$$\vec p\cdot\vec q = \frac19(-2-2+1)=-\frac13$$
$$\vec r\times\vec q = \frac19\begin{vmatrix}\hat i&\hat j&\hat k\-1&-1&2\-1&2&-1\end{vmatrix} = \frac19(3,3,3) = \frac{\hat i+\hat j+\hat k}3 \implies |\vec r\times\vec q|^2=\frac13$$

$$3\left(\frac13\right)^2-\lambda\left(\frac13\right)=0 \implies \frac13-\frac\lambda3=0 \implies \lambda=1$$

Solutions · 1