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Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 23 Maths Textbook Solution.

Answers (1)

Answer: $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

Hint: Use the properties of plane

Given: $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=5 \text { and } \overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=6$

Solution: We know that, the equation of line passing through $\left ( 1,2,3 \right )$ is given by

$\frac{x-1}{a_{1}}=\frac{y-2}{b_{1}}=\frac{z-3}{c_{1}}$ ………….. (1)

We know that line

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ if

$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …………… (2)

Here line (1) is parallel to plane

$\begin{gathered} x-2 y+z=5 \\ \end{gathered}$

So, $a \times 1+b \times(-1)+c \times 2=0 \\$

$\Rightarrow a-b+2 c=0$ …………... (3)

Also, line (1) parallel to plane

$\begin{gathered} 3 x+y+z=6 \\ \end{gathered}$

So, $a \times 3+b \times(1)+c \times 1=0 \\$

$\Rightarrow 3 a+b+c=0$ …………… (4)

Solving equation (3) and (4) by cross multiplication we have,

$\begin{aligned} &\frac{a}{-1 \times 1-(1) \times 2}=\frac{b}{3 \times 2-1 \times 1}=\frac{c}{1 \times 1-3 \times(-1)} \\ & \end{aligned}$

$\Rightarrow \frac{a}{-1-2}=\frac{b}{6-1}=\frac{c}{1+3} \\$

$\Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=k(\text { say })$

$a=-3 k, b=5 k \& c=4 k$

Putting the value in equation (1) we get

$\frac{x-1}{-3 k}=\frac{y-2}{5 k}=\frac{z-3}{4 k}$

Multiplying by k we have

$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$

The required equation is

$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})$

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