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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the vector equation of the plane through the points (2,1,-1) and (-1,3,4) and perpendicular to the plane x-2y+4z=10.

(Note: This question has been split into 2 questions)

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

Can you answer this question?
 
 

1 Answer

–1 vote
$\textbf{Step 1}$:
The required plane passes through: $P(2,1,-1)$ and $Q(-1,3,4)$.
Then we can write the position vectors as follows: $a_1=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ and $a_2=-\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}$
$\Rightarrow PQ=(a_2-a_1)=-3\overrightarrow{i}+2\overrightarrow{j}+5\overrightarrow{k}$
$\textbf{Step 2}$:
Let $\overrightarrow{i}$ be the normal vector to the desired plane $\overrightarrow{n}=\overrightarrow{n_i} \times \overrightarrow{PQ}$ =$\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & -2 & 4 \\ -3 & 2 & 5 \end{vmatrix}$
$\Rightarrow \overrightarrow n=\overrightarrow{i}(-10-8)-\overrightarrow{j}(5+12)+\overrightarrow{k}(2-6) = -18\overrightarrow{i}-17 \overrightarrow{j}-4 \overrightarrow{k}$
$\textbf{Step 3}$:
$\Rightarrow \overrightarrow{r}.\overrightarrow{n} =\overrightarrow{a}.\overrightarrow{n}$
$ \Rightarrow \overrightarrow r.(-18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k}) = (2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})(-18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k})$
$\Rightarrow$ $\overrightarrow{r}.(-18\overrightarrow{i}-17\overrightarrow{j}-4\overrightarrow{k})=-36-17+4$
$\Rightarrow$ $\overrightarrow{r}.(18\overrightarrow{i}-17\overrightarrow{j}+4\overrightarrow{k})=49$
answered Mar 21, 2013 by meena.p
edited Mar 22, 2013 by balaji.thirumalai
 

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