Get Answers to all your Questions

header-bg qa

Explain Solution R.D.Sharma Class 12 Chapter 28 The Plane  Exercise 28.13 Question 17 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:  a=1,4,5

Hint: use vector cross product

Given: x=5,\frac{y}{3-2}=\frac{z}{-2}  and x=\alpha ;\frac{y}{-1}=\frac{z}{2-\alpha }

Solution: we know that the lines

               \begin{aligned} &\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}\\ &\text { and }\\ \end{aligned}are coplar of

              \begin{aligned} &\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}} \end{aligned}

             \begin{aligned} &\left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0\\ \end{aligned}

            \begin{aligned} &\text { here }\\ &x_{1}=5, x_{2}=2, y_{1}=0, y_{2}\\ &=0, z_{1}=0, z_{2}=0\\ &l_{1}=1, l_{2}=1, m_{1}=3-a\\ &m_{2}=-1, n_{1}=-2, n_{2}=2-\alpha \end{aligned}

           \begin{gathered} \left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0 \\ \left(\begin{array}{ccc} \alpha-5 & 0 & 0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha \end{array}\right)=0 \end{gathered}

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads