Question

Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2)बिन्दुओं (2,1,0), (6,3,3) तथा (5,2,2) को समाहित करने वाले समतल में बिन्दु (2,1, 6) का प्रतिबिम्ब है-(10 (6,5,-2)(2) (6,-5,2)(3) (2,-3,4)(4) (2,-5, 6)​

Answer 1

Step-by-step explanation:

Given Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2)

• Let P be the point (2,1,0), Q be (6,3,3), and R be the point (5,2,2)
• Now the cross product of normal vector will be
• So PQ X PR
• So vector of PQ = 4i + 2j + 3k
• So vector     PR = 3i + j + 2k
• So we can write this in determinant, so we get
•                                i            j           k
•                                4           2          3
•                                3           1          2
• So PQ x PR = (4 – 3) i – j (8 – 9) + k (4 – 6)
•                   = i + j – 2k
• Now we need to find the equation of the plane. So equation of the plane will be
•       So x + y – 2z = C
• So (2,1,0) lies on this plane, so we get
•            2+ 1 – 0 = C
•             Or C = 3
• So the equation of plane will be x + y – 2z = 3
• Now in the point (2,1,6) we have an image.
• So image formula for P(x,y,z) will be  
• x – x1 / a = y – y1 / b = z – z1/c = - 2(ax1 + by1 + cz1 + d / a^2 + b^2 + c^2)
• x – 2 / 1 = y – 1 / 1 = z – 6 / - 2 = - 2(2 + 1 – 12 – 3 / 1 + 1 + 4)
•                                                   = - 2(- 12 / 6)
•                                                   = 4
• So x = 4 + 2 = 6
• So y = 4 + 1 = 5
• So z = (-8) + 6 = - 2
• So the coordinates of image will be (6, 5, - 2)

Reference link will be

https://brainly.in/question/8399515

Answer 2

Answer:

Step-by-step explanation:

Given Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2)

Let P be the point (2,1,0), Q be (6,3,3), and R be the point (5,2,2)

Now the cross product of normal vector will be

So PQ X PR

So vector of PQ = 4i + 2j + 3k

So vector     PR = 3i + j + 2k

So we can write this in determinant, so we get

                               i            j           k

                               4           2          3

                               3           1          2

So PQ x PR = (4 – 3) i – j (8 – 9) + k (4 – 6)

                  = i + j – 2k

Now we need to find the equation of the plane. So equation of the plane will be

      So x + y – 2z = C

So (2,1,0) lies on this plane, so we get

           2+ 1 – 0 = C

            Or C = 3

So the equation of plane will be x + y – 2z = 3

Now in the point (2,1,6) we have an image.

So image formula for P(x,y,z) will be  

x – x1 / a = y – y1 / b = z – z1/c = - 2(ax1 + by1 + cz1 + d / a^2 + b^2 + c^2)

x – 2 / 1 = y – 1 / 1 = z – 6 / - 2 = - 2(2 + 1 – 12 – 3 / 1 + 1 + 4)

                                                  = - 2(- 12 / 6)

                                                  = 4

So x = 4 + 2 = 6

So y = 4 + 1 = 5

So z = (-8) + 6 = - 2

So the coordinates of image will be (6, 5, - 2)