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  • Need clarity, kindly explain! The sum of the intercepts on the coordinate axes of the plane passing through the point(−2, −2, 2) and containing the line joining the points (1, −1, 2) and (1, 1, 1),is :

The sum of the intercepts on the coordinate axes of the plane passing through the point
(−2, −2, 2) and containing the line joining the points (1, −1, 2) and (1, 1, 1),
is :

  • Option 1)

    4

  • Option 2)

    -4

  • Option 3)

    -8

  • Option 4)

    12

 

Answers (1)

best_answer

As we have learned

Cartesian equation of plane passing through a given point and normal to a given vector -

\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0
 

- wherein

\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}

\vec{a}= x_{0}\hat{i}+y_{0}\hat{j}+z_{0}\hat{k}

\vec{n}= a\hat{i}+b\hat{j}+c\hat{k}

Putting in

\left ( \vec{r}-\vec{a} \right )\cdot \vec{n}= 0

We get \left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0

 

 

 

a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0

a(x+2)+b(y+2)+c(z-2)=0

passing through (1,-1,2)

 

 and (1,1,1)

\Rightarrow 3b+b+0c=0

\Rightarrow 3b+3b-c=0

\frac{a}{-1}=\frac{b}{3}=\frac{c}{6}

-x-2+3y+6+62-12=0

-x+3y+6z=8

\frac{x}{-8}+\frac{y}{8/3}+\frac{z}{4/3}=1

sum of intercepts = -4

 

 

 

 


Option 1)

4

This is incorrect

Option 2)

-4

This is correct

Option 3)

-8

This is incorrect

Option 4)

12

This is incorrect

Posted by

Himanshu

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