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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 11 Sub Question 1 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 11 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer p(1,1,-2)

                Hint: use vector cross product i+j-2k,2i-j+k   and i+2j+k 

Solution: let A, B and C be three point with position vector

              i+j-2k,2i-j+k   andi+2j+k

Thus  \begin{aligned} &\overrightarrow{A B}=\vec{b}-\vec{a}=(2 i-j+k)-(i+j-2 k) \\ \end{aligned}

          \begin{aligned} &=i-2 j+3 k \\ &\overrightarrow{A C}=\vec{c}-\vec{a}=(i+2 j+k)-(i+j-2 k)=j+3 k \end{aligned}

As we know that cross product of the line vectors gives a perpendicular vector so

                   \begin{aligned} &\vec{m}=\overrightarrow{A B} x \overrightarrow{A C}=\left(\begin{array}{ccc} i & j & k \\ 1 & -2 & 3 \\ 0 & 1 & 3 \end{array}\right) \\ &\vec{m}=i(-6-3)-3 j+k=9 i-3 j+k \end{aligned}

So the equation of the required plane is

                \begin{aligned} &(\vec{r}-\vec{a}) \vec{n}=0 \\ &(\vec{r}-\vec{n}) \cdot \vec{n}=(\vec{a}-\vec{n}) \\ &(\vec{r} \cdot(-9 i-3 j+k))=(i+j-2 k)-(-9 i-3 i+k) \\ &\vec{r}=(-9 i-3 j+k)=14 \end{aligned}

Also we have to find the coordinates of the point of intersection of the plane and the line

\vec{r}=3i-j-k+\lambda \left ( 2i-2j+k \right )   any point on the line

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