Home
Class 12
PHYSICS
In a parallelogram ABCD. Prove that vec(...

In a parallelogram ABCD. Prove that `vec(AC)+ vec (BD) = 2 vec(BC)`

Text Solution

AI Generated Solution

The correct Answer is:
To prove that in a parallelogram ABCD, \( \vec{AC} + \vec{BD} = 2 \vec{BC} \), we will follow these steps: ### Step 1: Understand the structure of the parallelogram In a parallelogram ABCD, the points A, B, C, and D are the vertices such that: - \( \vec{AB} \) is parallel to \( \vec{CD} \) - \( \vec{AD} \) is parallel to \( \vec{BC} \) ### Step 2: Express the diagonal vectors The diagonals of the parallelogram can be expressed in terms of the sides: - The vector \( \vec{AC} \) can be expressed as: \[ \vec{AC} = \vec{AB} + \vec{BC} \] - The vector \( \vec{BD} \) can be expressed as: \[ \vec{BD} = \vec{BA} + \vec{AD} \] Since \( \vec{BA} = -\vec{AB} \) (because \( \vec{BA} \) is in the opposite direction of \( \vec{AB} \)), we can rewrite \( \vec{BD} \) as: \[ \vec{BD} = -\vec{AB} + \vec{AD} \] ### Step 3: Use the properties of the parallelogram Since \( \vec{AD} \) is equal to \( \vec{BC} \) (as opposite sides of a parallelogram are equal), we can substitute \( \vec{AD} \) with \( \vec{BC} \): \[ \vec{BD} = -\vec{AB} + \vec{BC} \] ### Step 4: Add the two diagonal vectors Now, we can add \( \vec{AC} \) and \( \vec{BD} \): \[ \vec{AC} + \vec{BD} = (\vec{AB} + \vec{BC}) + (-\vec{AB} + \vec{BC}) \] ### Step 5: Simplify the expression When we simplify the expression, we find: \[ \vec{AC} + \vec{BD} = \vec{AB} - \vec{AB} + \vec{BC} + \vec{BC} = 0 + 2\vec{BC} = 2\vec{BC} \] ### Conclusion Thus, we have proven that: \[ \vec{AC} + \vec{BD} = 2\vec{BC} \]
To prove that in a parallelogram ABCD, \( \vec{AC} + \vec{BD} = 2 \vec{BC} \), we will follow these steps: ### Step 1: Understand the structure of the parallelogram In a parallelogram ABCD, the points A, B, C, and D are the vertices such that: - \( \vec{AB} \) is parallel to \( \vec{CD} \) - \( \vec{AD} \) is parallel to \( \vec{BC} \) ### Step 2: Express the diagonal vectors ...
Promotional Banner

Similar Questions

Explore conceptually related problems

vec(AC) and vec(BD) are the diagonals of a parallelogram ABCD. Prove that (i) vec(AC) + vec(BD) - 2 vec(BC) (ii) vec(AC) - vec(BD) - 2vec(AB)

Given a parallelogram ABCD . If |vec(AB)|=a, |vec(AD)| = b & |vec(AC)| = c , then vec(DB) . vec(AB) has the value

If ABCD is a parallelogram, then vec(AC) - vec(BD) =

E and F are the interior points on the sides BC and CD of a parallelogram ABCD. Let vec(BE)=4vec(EC) and vec(CF)=4vec(FD) . If the line EF meets the diagonal AC in G, then vec(AG)=lambda vec(AC) , where lambda is equal to :

Let B_1,C_1 and D_1 are points on AB,AC and AD of the parallelogram ABCD, such that vec(AB_1)=k_1vec(AC,) vec(AC_1)=k_2vec(AC) and vec(AD_1)=k_2 vec(AD,) where k_1,k_2 and k_3 are scalar.

Prove that the area of a parallelogram with sides vec(A) and vec(B) " is " vec(A) xx vec(B)

In a regular hexagon ABCDEF, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

In a quadrilateral ABCD, vec(AB) + vec(DC) =

ABCD is a parallelogram with vec(AC) = hati - 2hatj + hatk and vec(BD) = -hati + 2hatj - 5hatk . Area of this parallelogram is equal to:

Let ABCD be a parallelogram such that vec A B= vec q , vec A D= vec p""a n d""/_B A D be an acute angle. If vec r is the vector that coincides with the altitude directed from the vertex B to the side AD, then vec r is given by (1) vec r=3 vec q-(3( vec pdot vec q))/(( vec pdot vec p)) vec p (2) vec r=- vec q+(( vec pdot vec q)/( vec pdot vec p)) vec p (3) vec r= vec q+(( vec pdot vec q)/( vec pdot vec p)) vec p (4) vec r=-3 vec q+(3( vec pdot vec q))/(( vec pdot vec p)) vec p