A plane containing the point (3, 2, 0) and the line  \frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}  also contains the point :

  • Option 1)

    (0, -3, 1)

  • Option 2)

    (0, 7, 10)

  • Option 3)

    (0, 7, -10)

  • Option 4)

    (0, 3, 1)

 

Answers (1)

As we learnt in 

Normal form (cartesian form ) -

lx+my+nz=d

where d is the distance from origin.  

- wherein

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

\hat{n}= l\hat{i}+m\hat{j}+n\hat{k}

putting in \vec{r}\cdot \hat{n}= d

We get -  lx+my+nz= d

 

 Normal vector of Plane is =\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 3-1& 2-2& 0-3\\ 1 & 5 & 4 \end{vmatrix}

 

=\begin{vmatrix} \hat{i} & \hat{j}& \hat{k}\\ 2 & 0& -3\\ 1 & 5 & 4 \end{vmatrix}=15\hat{i}-11\hat{j}+10\hat{k}

15x-11y+10z=23

It passes through (0,7,10)


Option 1)

(0, -3, 1)

This option is incorrect

Option 2)

(0, 7, 10)

This option is correct

Option 3)

(0, 7, -10)

This option is incorrect

Option 4)

(0, 3, 1)

This option is incorrect

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