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  • Stuck here, help me understand: A plane bisects the line segment joining the points (1, 2, 3) and (−3, 4, 5) at right angles. Then this plane also passes through the point :

A plane bisects the line segment joining the points (1, 2, 3) and (−3, 4, 5) at right angles. Then this plane also passes through the point :

  • Option 1)

    (−3, 2, 1)

  • Option 2)

    (3, 2, 1)

  • Option 3)

    (−1, 2, 3)

  • Option 4)

    (1, 2, −3)

 

Answers (1)

best_answer

As we 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

 

 and

 

Direction Ratios -

(i)     if a,b,c are direction ratios then direction cosines will be

l=\frac{\pm a}{\sqrt{a^{2}+b^{2}+c^{2}}},m=\frac{\pm b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{\pm c}{\sqrt{a^{2}+b^{2}+c^{2}}}

(ii)        Direction ratios of line joining two given points

A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right ) is given by

        \left ( x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1} \right )

(iii)    If  r= a\hat{i}+b\hat{j}+c\hat{k} be a vector with direction cosines l, m, n then

l= \frac{a}{\left | r \right |},m= \frac{b}{\left | r \right |},n= \frac{c}{\left | r \right |}

-

 

 

Plane must meet line at midpoint of line 

i.e.  (-1,3,4)

DCs of plane \Rightarrow (-4,2,2)

equation of plane -4x+2y+2z=\lambda

\Rightarrow -4x+2y+2z=18

Only (-3,2,1) satisfies it


Option 1)

(−3, 2, 1)

Option 2)

(3, 2, 1)

Option 3)

(−1, 2, 3)

Option 4)

(1, 2, −3)

Posted by

Himanshu

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