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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry1 Mark
MCQ
The eqution of the plane $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$ in scalar product from is:
  • ✓
    $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
  • B
    $\vec{\text{r}}.(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=7$
  • C
    $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})=7$
  • D
    None of these
✓

Answer

Correct option: A.
$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represents a plane passing through a point whose position vectors is $\vec{\text{a}}$ and parrallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}};\ \vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}} \hat{\text{j}} \hat{\text{k}}1 1 11 -2 3\end{vmatrix}$
$=5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product from is
$\vec{\text{r}}.\vec{\text{n}}=\vec{\text{a}}.\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(\hat{\text{i}}-\hat{\text{j}}-0\hat{\text{k}}\big).\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=5+2+0$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$

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