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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 5 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 5  maths textbook solution

Answers (1)

Answer:

\begin{aligned} &\vec{r}=(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+\lambda(\hat{\imath}-13 \hat{\jmath}+17 \hat{k}) \\ &\frac{x-2}{1}=\frac{y+3}{13}=\frac{z-4}{17} \end{aligned}

Hint: Position vector of the mid point

Given : ABCD is a Parallelogram

Solution:

We know that the position vector of the mid-point of \begin{aligned} &\vec{a} \end{aligned} and is \begin{aligned} &\vec{b} \end{aligned}is \frac{\vec{a}+\vec{b}}{2} 

Let the position vector of D be x\hat{i}+y\hat{j}+z\hat{k}

 Position vector of mid-point of A and C =  Position vector of mid-point of B and D

\begin{aligned} &\frac{(4 \hat{\mathrm{i}}+5 \hat{\mathbf{j}}-10 \hat{\mathrm{k}})+(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{2}=\frac{(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(x\hat{\mathrm{i}}+ y\hat{\mathbf{j}}+z\hat{\mathrm{k}})}{2} \\ &\Rightarrow \frac{3}{2} \hat{\mathrm{i}}+\frac{7}{2} \hat{\mathrm{j}}-\frac{9}{2} \hat{\mathrm{k}}=\left(\frac{\mathrm{x}+2}{2}\right) \hat{\mathrm{i}}+\left(\frac{-3+\mathrm{y}}{2}\right) \hat{\mathrm{j}}+\left(\frac{4+\mathrm{z}}{2}\right) \hat{\mathrm{k}} \end{aligned}

Comparing the coefficients of \hat{i},\hat{j} & \hat{k} we get

              \begin{aligned} &\left(\frac{x+2}{2}\right)=\frac{3}{2} \\ &\Rightarrow x=1 \\ &\frac{-3+y}{2}=\frac{7}{2} \\ &\Rightarrow y=10 \\ &\frac{4+z}{2}=-\frac{9}{2} \\ &\Rightarrow z=-13 \end{aligned}

Position vector of point D=\hat{i}+10\hat{j}-13\hat{k}. The vector equation of line BD passing through the points with position vectors

               \overrightarrow{\mathrm{a}}(\mathrm{B}) and \overrightarrow{\mathrm{b}}(\mathrm{D})

               \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})

                 Here,

                \begin{aligned} &\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}} \\ &\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+10 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \end{aligned}

Vector equation of the required line is

            \begin{aligned} &\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}})+\lambda\{(\hat{\mathrm{i}}+10 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})-(2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}})\} \\ &\Rightarrow \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+\lambda(-\hat{\mathrm{i}}+13 \hat{\mathbf{j}}-17 \hat{\mathrm{k}}) \end{aligned}  ………….. (1)

Here, \lambdais a parameter.

Reducing (1) to Cartesian form, we get

\begin{aligned} &x \hat{\mathbf{i}}+y \hat{\mathrm{j}}+\mathrm{z} \hat{\mathbf{k}}=(2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}})+\lambda(-\hat{\mathrm{i}}+13 \hat{\mathrm{j}}-17 \hat{\mathrm{k}})[\text { Putting } \overrightarrow{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}} \text { in }(1)] \\ &\Rightarrow \mathrm{x} \hat{\mathbf{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}=(2-\lambda) \hat{\mathrm{i}}+(-3+13 \lambda) \hat{\mathrm{j}}+(4-17 \lambda) \hat{\mathrm{k}} \end{aligned}

Comparing the coefficients of \hat{i},\hat{j}  & \hat{k} we get

\begin{aligned} &x=2-\lambda, y=-3+13 \lambda, \mathrm{z}=4-17 \lambda \\ &\Rightarrow \frac{x-2}{-1}=\lambda, \frac{y+3}{13}=\lambda, \frac{z-4}{-17}=\lambda \\ &\Rightarrow \frac{x-2}{-1}=\frac{y+3}{13}=\frac{z-4}{-17}=\lambda \\ &\Rightarrow \frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}=-\lambda \end{aligned}

Hence, the Cartesian form of (1) is

\frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}

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