In ∆ABC,D and E divides AB and AC in the ratio of 1:2 respectively. Prove that Vector DE=1/3BC.
Make a sketch
If D divides AB in the ratio of 1 : 2,
then DA = 1/3 BA
similarly EA = 1/3 CA
notice I was careful in the order I wrote EA, that way writing them as vectors would be the same.
I will use the notation v(AB) for vector AB
v(DE) = v(DA) + v(AE)
= (1/3)v(BA) + (1/3)v(AC)
= (1/3)( v(BA) + v(AC) )
= 1/3 v(BC)
as required
Well, it seems like this problem has got its ratios all mixed up! It's trying to divide things left and right. Let me give it a shot and sprinkle some humor along the way!
So, we have triangle ABC, right? And we have points D and E dividing sides AB and AC in a 1:2 ratio, respectively. We want to prove that vector DE is equal to 1/3 the length of vector BC.
Now, before we start, let's make sure D and E are not playing any funny business here. If D and E are dividing AB and AC in that ratio, they better not be throwing any wild parties or going on extreme adventures! Otherwise, our calculation might be way off!
To prove that vector DE is equal to 1/3 of vector BC, let's look at the ratios involved. D divides AB in a 1:2 ratio, so we can say AD is 1/3 of AB, and DB is 2/3 of AB. Similarly, E divides AC in a 1:2 ratio, so AE is 1/3 of AC, and EC is 2/3 of AC.
Now, let's look at vector DE. We can express vector DE as vector AD + vector AE. Since AD is 1/3 of AB, we can write it as (1/3)*AB, and AE, being 1/3 of AC, can be expressed as (1/3)*AC.
Now, let's substitute these values into our equation. We have vector DE = (1/3)*AB + (1/3)*AC. And guess what? We can factor out that 1/3, which is like saying we're dividing by 3!
So, our equation becomes vector DE = 1/3 * (AB + AC). And wait a minute, that AB + AC looks awfully familiar! That's actually vector BC! So, we can rewrite our equation as vector DE = 1/3 * BC.
Tada! We've proved that vector DE is indeed equal to 1/3 of vector BC! Don't you just love it when calculations add up nicely like this? Now, let's hope D and E won't go on any more adventures and throw off our math game!
To prove that vector DE is equal to 1/3 times vector BC, we can use the concept of position vectors.
Let's assume that the position vector of point A is vector a, and the position vectors of points B and C are vector b and vector c, respectively. Therefore, the position vectors of points D and E can be written as:
Position Vector of D: vector d = (1/3) * vector b
Position Vector of E: vector e = (2/3) * vector c
Now, we can calculate vector DE by subtracting vector d from vector e:
vector DE = vector e - vector d
= (2/3) * vector c - (1/3) * vector b
To prove that vector DE is equal to 1/3 times vector BC, we need to show that:
vector DE = (1/3) * vector BC
Since point B has position vector vector b and point C has position vector vector c, we can write vector BC as:
vector BC = vector c - vector b
Therefore, to prove the statement, we need to show that:
(2/3) * vector c - (1/3) * vector b = (1/3) * (vector c - vector b)
Let's simplify both sides of the equation:
(2/3) * vector c - (1/3) * vector b = (1/3) * (vector c - vector b)
(2/3) * vector c - (1/3) * vector b = (1/3) * vector c - (1/3) * vector b
Multiplying each term by 3 to eliminate the fractions:
2 * vector c - vector b = vector c - vector b
Now, we can see that vector DE is equal to 1/3 times vector BC, as both sides of the equation are equal.
Hence, vector DE = (1/3) * vector BC.
To prove that vector DE is equal to 1/3 * vector BC, we need to establish an equality between the two vectors.
Let's begin by considering the ratio in which D divides AB and E divides AC.
Given that D divides AB in the ratio 1:2, we can represent the position vector of point D as:
Vector AD = 1/3 * vector AB
Similarly, since E divides AC in the ratio 1:2, we can represent the position vector of point E as:
Vector AE = 2/3 * vector AC
Now, let's calculate the position vector of point D relative to point E. This can be done by subtracting the position vector of E from the position vector of D:
Vector DE = Vector AD - Vector AE
Substituting the values of vector AD and vector AE derived earlier, we get:
Vector DE = (1/3 * vector AB) - (2/3 * vector AC)
Now, we can manipulate this equation to simplify the expression:
Vector DE = (1/3 * vector AB) + (-2/3 * vector AC)
Rewriting the equation by taking out the common factor of 1/3:
Vector DE = 1/3 * (vector AB + (-2 * vector AC))
Next, distributing the scalar, -2, to vector AC:
Vector DE = 1/3 * (vector AB - 2 * vector AC)
Now, notice that vector AB - 2 * vector AC is equal to vector BC, as it follows the laws of vector addition and subtraction.
Therefore, substituting in the equation:
Vector DE = 1/3 * vector BC
And thus, we have successfully proven that vector DE is equal to 1/3 * vector BC.