A plane meets the coordinate axes in A , B , C such that the centroid of the triangle A B C is the point p , q , r . Show that the equation of the plane is x p + y q + z r = 3

Given centroid of the triangle A B C is ( p , q , r ) Let the required equation of the plane is x a + y b + z c = 1 . . . . 1 where a , b , c , are x - intercept, y - intercept, z - intercept respectively. Thus, coordinates of A , where the plane meets the x -axis are a , 0 , 0 , coordinates of B are 0 , b , 0 , and coordinates of C where the plane cuts the z -axis are 0 , 0 , c Centroid of the triangle A B C is ( p , q , r ) = a + 0 + 0 3 , 0 + b + 0 3 , 0 + 0 + c 3 ⇒ ( p , q , r ) = a 3 , b 3 , c 3 . We know that two ordered pairs are equal if and only if the corresponding components of the two are equal. So, a = 3 p , b = 3 q and c = 3 r Now, substitute these values in equation 1 ⇒ x 3 p + y 3 q + z 3 r = 1 Hence, equation of the plane is x p + y q + z r = 3 .

If α , β , γ are direction angles of the vector O P → , then prove that cos 2 α + cos 2 β + cos 2 γ = 1 . Hence, deduce that sin 2 α + sin 2 β + sin 2 γ = 2

Consider the three-dimensional right-handed rectangular coordinate system. Consider a point P in space, having coordinates x , y , z with respect to the origin O 0 , 0 , 0 . Then, the magnitude of vector O P is given by | O P → | = x 2 + y 2 + z 2 . . . . i Now, it is given that The angles α , β , γ made by the O P → (or r → ) with the positive directions of x , y , and z -axes respectively, are called its direction angles. The cosine values of these angles, i.e., cos α , cos β , cos γ are called direction cosines of the vector r → . From the above figure, from triangle O A P , we have cos α = x r From the right angled triangle O B P , we have cos β = y r And, from the right-angled triangle O C P , we have cos γ = z r Now, cos 2 α + cos 2 β + cos 2 γ = x 2 + y 2 + z 2 r 2 = O P → 2 r 2 [Using i ] = r 2 r 2 = 1 Hence, Proved. Now, consider cos 2 α + cos 2 β + cos 2 γ = 1 ⇒ 1 - sin 2 α + 1 - sin 2 β + 1 - sin 2 γ = 1 ⇒ sin 2 α + sin 2 β + sin 2 γ = 3 - 1 = 2 Hence, Proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Three Dimensional Geometry>Maharastra Board-2018

Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Maharastra Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Maharastra Board-2018

Attempt the free practice questions on Chapter 6: Three Dimensional Geometry, Exercise 1: Maharastra Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Maharastra Board-2018 with Hints & Solutions

MEDIUM

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Three Dimensional Geometry>Maharastra Board-2019>Q 1

Show that the points $A (- 7, 4, - 2), B (- 2, 1, 0)$ and $C (3, - 2, 2)$ are collinear.

MEDIUM

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Three Dimensional Geometry>Maharastra Board-2018>Q 2

A plane meets the coordinate axes in $A, B, C$ such that the centroid of the triangle $A B C$ is the point $= \sqrt{25 + 9 + 4}$ . Show that the equation of the plane is $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 3$

HARD

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Three Dimensional Geometry>Maharastra Board-2018>Q 3

If $α, β, γ$ are direction angles of the vector $= \sqrt{100 + 36 + 16}$ , then prove that $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$ . Hence, deduce that $\sin^{2} α + \sin^{2} β + \sin^{2} γ = 2$