Get Answers to all your Questions

header-bg qa

If the point (2,\alpha ,\beta ) lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane 2x-5y=15 ,

then 2\alpha -3\beta is equal to :

  • Option 1)

    17

  • Option 2)

    5

  • Option 3)

    7

  • Option 4)

    12

Answers (1)

best_answer

 

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

 

 

Conversion of equation in normal form (vector form ) -

The equation ax+by+cz+D= 0

is converted in normal by

(i)    ax+by+cz= -D

(ii)    Making RHS position

    -ax-by-cz= D

(iii)    Dividing by \sqrt{a^{2}+b^{2}+c^{2}}

(iv)    \frac{-ax-by-cz}{\sqrt{a^{2}+b^{2}+c^{2}}}= \frac{D}{\sqrt{a^{2}+b^{2}+c^{2}}}

We get lx+my+nz= d

-

 

Normal Vector of plane =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2& -5 & 0\\ 4 & -4 & 4 \end{vmatrix}=-4(5\hat{i}+2\hat{j}-3\hat{k})

Equation of plane is 5(x-7)+2y-3(z-6)=0

=>5x+2y-3z=17

=>2\alpha -3\beta =17-(5\times 2)=7


Option 1)

17

Option 2)

5

Option 3)

7

Option 4)

12

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE