Solveeit Logo
Login

Question

Mathematics Question on Three Dimensional Geometry

The plane r→=s(i^+2j^−4k^)+t(3i^+4j^−4k^)\overrightarrow{r}=s(\hat{i}+2\hat{j}-4\hat{k})+t(3\hat{i}+4\hat{j}-4\hat{k}) +(1−t)(2i^−7j^−3k^)+(1-t)(2\hat{i}-7\hat{j}-3\hat{k}) is parallel to the line

A

r→=(−i^+j^−k^)+t(−i^−2j^+4k^)\overrightarrow{r}=(-\hat{i}+\hat{j}-\hat{k})+t(-\hat{i}-2\hat{j}+4\hat{k})

B

r→=(−i^+j^−k^)+t(i^−2j^+4k^)\overrightarrow{r}=(-\hat{i}+\hat{j}-\hat{k})+t(\hat{i}-2\hat{j}+4\hat{k})

C

r→=(i^+j^−k^)+t(−i^−4j^+7k^)\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+t(-\hat{i}-4\hat{j}+7\hat{k})

D

r→=(−i^+j^−k^)+t(−2i^+2j^+4k^)\overrightarrow{r}=(-\hat{i}+\hat{j}-\hat{k})+t(-2\hat{i}+2\hat{j}+4\hat{k})

Answer

r→=(−i^+j^−k^)+t(−i^−2j^+4k^)\overrightarrow{r}=(-\hat{i}+\hat{j}-\hat{k})+t(-\hat{i}-2\hat{j}+4\hat{k})

Explanation

Solution

Given plane is
⇒\Rightarrow r→=r(i^+2j^−4k^)+t(3i^+4j^−4k^)\overrightarrow{r}=r(\hat{i}+2\hat{j}-4\hat{k})+t(3\hat{i}+4\hat{j}-4\hat{k}) +(1−t)(2i^−7j^−3k^)+(1-t)(2\hat{i}-7\hat{j}-3\hat{k})
⇒\Rightarrow r→=(2i^−7j^+3k^)+s(i^+2j^−4k^)\overrightarrow{r}=(2\hat{i}-7\hat{j}+3\hat{k})+s(\hat{i}+2\hat{j}-4\hat{k}) +t(i^+11j^−k^)+t(\hat{i}+11\hat{j}-\hat{k})
Comparing it with the equation of plane r→=a→+λb→+μc→,\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}+\mu \overrightarrow{c}, we get b→=i^+2j^−4k^\overrightarrow{b}=\hat{i}+2\hat{j}-4\hat{k} and c→=i^+11j^−k^\overrightarrow{c}=\hat{i}+11\hat{j}-\hat{k}
Now, b→×c→=∣i^j^k^ 12−4 111−1 ∣\overrightarrow{b}\times \overrightarrow{c}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 2 & -4 \\\ 1 & 11 & -1 \\\ \end{matrix} \right|
=42i^−3j^+9k^=42\hat{i}-3\hat{j}+9\hat{k}
∴\therefore Parametric form of plane is r→.(b→×c→)=a→.(b→×c→)\overrightarrow{r}.(\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})
⇒\Rightarrow r→.(42i^−3j^+9k^)\overrightarrow{r}.(42\hat{i}-3\hat{j}+9\hat{k})
=(2i^−7j^+3k^).(42i^−3j^+9k^)=(2\hat{i}-7\hat{j}+3\hat{k}).(42\hat{i}-3\hat{j}+9\hat{k})
which is of the form r→.r→=d\overrightarrow{r}.\overrightarrow{r}=d
⇒\Rightarrow r→=42i^−3j^+9k^\overrightarrow{r}=42\hat{i}-3\hat{j}+9\hat{k}
Now, the line given in option (a) is
r→=(−i^+j^+k^)+t(−i^−2j^+4k^)\overrightarrow{r}=(-\hat{i}+\hat{j}+\hat{k})+t(-\hat{i}-2\hat{j}+4\hat{k})
Comparing it with r→=p→+tq→,\overrightarrow{r}=\overrightarrow{p}+t\overrightarrow{q}, we get q→=(−i^−2j^+4k^)\overrightarrow{q}=(-\hat{i}-2\hat{j}+4\hat{k}) Since, q→.n→=(−i^−2j^+4k^).(42i^−3j^+9k^)\overrightarrow{q}.\overrightarrow{n}=(-\hat{i}-2\hat{j}+4\hat{k}).(42\hat{i}-3\hat{j}+9\hat{k})
=−42+6+36=0=-42+6+36=0
Hence, the line given in option (a) is parallel to the given plane.