Let _1,_2 be the values of for which the points (52,1,) and (-2,0,1) are at equal distance…

Question

Let $\lambda_1,\lambda_2$ be the values of $\lambda$ for which the points $\left(\dfrac52,1,\lambda\right)$ and $(-2,0,1)$ are at equal distance from the plane $2x+3y-6z+7=0$. If $\lambda_1>\lambda_2$, then the distance of the point $(\lambda_1-\lambda_2,\ \lambda_2,\ \lambda_1)$ from the line $\dfrac{x-5}{1}=\dfrac{y-1}{2}=\dfrac{z+7}{2}$ is:

Answer: 9

Accepted Answer

$$d_1=\frac{|5+3-6\lambda+7|}{7}=\frac{|15-6\lambda|}{7},\qquad d_2=\frac{|-4-6+7|}{7}=\frac{3}{7}.$$
$$|15-6\lambda|=3 \implies \lambda=2\text{ or }3.$$
So $\lambda_1=3,\ \lambda_2=2$, and $P=(1,2,3)$.

Line point $(5,1,-7)$, direction $(1,2,2)$; $\vec{(\text{point})P}=(-4,1,10)$.
$$\vec{(\text{point})P}\times(1,2,2)=(-18,18,-9),\qquad |(-18,18,-9)|=27.$$
$$\text{Distance}=\frac{27}{3}=9.$$

Solutions · 1