CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let the line $$\displaystyle \frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2}$$ lie in the plane $$x+3y-\alpha z+\beta = 0$$. Then $$\left ( \alpha ,\beta  \right )$$ equals :


A
(6,7)
loader
B
(5,15)
loader
C
(5,5)
loader
D
(6,17)
loader

Solution

The correct option is A $$\left ( -6,7 \right )$$
The line is $$\displaystyle \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}2{}$$

The direction ratios of the line are $$(3,-5,2)$$

As the line is in the plane $$x+3y-az+ \beta =0$$,

We have $$\left ( 3 \right )\left ( 1 \right )+\left ( -5 \right )\left ( 3 \right )+2\left ( -\alpha  \right )=0$$

$$\Rightarrow-12-2 \alpha =0$$

$$ \therefore \alpha = -6$$

Again $$(2,1,-2)$$ lies on the plane

$$\Rightarrow 2+3+2 \alpha + \beta =0$$

$$\Rightarrow \beta = -2 \alpha -5=12-5=7$$

Hence, $$\left ( \alpha ,\beta  \right )$$ is $$\left ( -6,7 \right )$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image