Home
Class 12
MATHS
Statement -1 If xbara+ybarb+zbarc=0 impl...

Statement -1 If `xbara+ybarb+zbarc=0 implies x+y+z=0` where x,y,z are scalars and `bara,barb,barc` are p.v of three points A,B,C then A,B,C are linearly dependent.
Statement-2: A,B,C are collinear points.

A

Statement -1 is true, Statement -2 is true, Statement-2 is a correct explanation for Statement -1

B

Statement-1 is true,Statement-2 is true,Statement-2is NOT a correct explanation for Statement-1

C

Statement-1 is True,Statement-2 is false

D

Statement-1 is false, Statement-2 is true

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will analyze the statements given and derive the necessary conclusions step by step. ### Step 1: Understanding the Given Information We have two statements: - **Statement 1**: If \( \bar{x} \cdot \vec{a} + \bar{y} \cdot \vec{b} + \bar{z} \cdot \vec{c} = 0 \) implies \( x + y + z = 0 \), where \( x, y, z \) are scalars and \( \vec{a}, \vec{b}, \vec{c} \) are position vectors of points A, B, and C respectively. - **Statement 2**: A, B, C are collinear points. ### Step 2: Analyzing Statement 1 From the equation \( \bar{x} \cdot \vec{a} + \bar{y} \cdot \vec{b} + \bar{z} \cdot \vec{c} = 0 \), we can interpret this as a linear combination of the vectors \( \vec{a}, \vec{b}, \vec{c} \) equating to zero. If \( x + y + z = 0 \), we can express one of the variables in terms of the others. For example, we can express \( z \) as: \[ z = -x - y \] ### Step 3: Substituting into the Vector Equation Substituting \( z \) into the vector equation gives: \[ \bar{x} \cdot \vec{a} + \bar{y} \cdot \vec{b} - (x + y) \cdot \vec{c} = 0 \] This implies that the vectors \( \vec{a}, \vec{b}, \vec{c} \) are linearly dependent because we can express one vector as a linear combination of the others. ### Step 4: Conclusion from Statement 1 Since the vectors are linearly dependent, we can conclude that points A, B, and C are linearly dependent. Therefore, Statement 1 is true. ### Step 5: Analyzing Statement 2 If points A, B, and C are collinear, it means they lie on the same straight line. Collinearity of points implies that they are linearly dependent, as one point can be expressed as a linear combination of the others. ### Step 6: Conclusion from Statement 2 Since Statement 2 states that A, B, and C are collinear points, and we have established that linear dependence implies collinearity, Statement 2 is also true. ### Final Conclusion Both statements are true: - **Statement 1**: True (A, B, C are linearly dependent) - **Statement 2**: True (A, B, C are collinear)
Promotional Banner

Similar Questions

Explore conceptually related problems

If x,y,z are in G.P and a^x=b^y=c^z ,then

Let a,b,c be in A.P and x,y,z be in G.P.. Then the points (a,x),(b,y) and (c,z) will be collinear if

IF bar(a),bar(b),bar(c ) are vectors from the origin to three points A,B,C show that bara times barb +barb times barc+barc times bara is perpendicular to the plane ABC.

If a^x=b^y=c^z and a,b,c are in G.P. show that 1/x,1/y,1/z are in A.P.

Consider the equation az + b barz + c =0 , where a,b,c in Z If |a| = |b| and barac ne b barc , then z has

Statement -1 : If a transversal cuts the sides OL, OM and diagonal ON of a parallelogram at A, B, C respectively, then (OL)/(OA) + (OM)/(OB) =(ON)/(OC) Statement -2 : Three points with position vectors veca , vec b , vec c are collinear iff there exist scalars x, y, z not all zero such that x vec a + y vec b +z vec c = vec 0, " where " x +y + z=0.

If O be the origin the vector vec(OP) is called the position vector of point P. Also vec(AB)=vec(OB)-vec(OA) . Three points are said to be collinear if they lie on the same stasighat line.Points A,B,C are collinear if one of them divides the line segment joining the others two in some ratio. Also points A,B,C are collinear if and only if vec(AB)xxvec(AC)=vec0 Let the points A,B, and C having position vectors veca,vecb and vecc be collinear Now answer the following queston: The exists scalars x,y,z such that (A) xveca+yvecb+zcvecc=0 and x+y+z!=0 (B) xveca+yvecb+zcvecc!=0 and x+y+z!=0 (C) xveca+yvecb+zvecc=0 and x+y+z=0 (D) none of these

Statement 1 : If the chords of contact of tangents from three points A ,B and C to the circle x^2+y^2=a^2 are concurrent, then A ,B and C will be collinear. Statement 2 : Lines (a_1x+b_1y+c_1)+k(a_2x+b_2y+c_2)=0 alwasy pass through a fixed point for k in R .

Let z_1, z_2, z_3 be three complex numbers and a ,b ,c be real numbers not all zero, such that a+b+c=0a n da z_1+b z_2+c z_3=0. Show that z_1, z_2,z_3 are collinear.

Let z_1, z_2, z_3 be three complex numbers and a ,b ,c be real numbers not all zero, such that a+b+c=0 and a z_1+b z_2+c z_3=0. Show that z_1, z_2,z_3 are collinear.