Vectors

Step 1: Define the Position Vector of $C$
Let

C

divide

BA

externally in the ratio

5%3A3

. The position vector of

C

can be expressed as:

%5Cmathbf%7Bc%7D%20%3D%20%5Cfrac%7B5%5Cmathbf%7Ba%7D%20-%203%5Cmathbf%7Bb%7D%7D%7B5%20-%203%7D.

Step 2: Use the Given Equation
Rearranging the given equation

5%5Cmathbf%7Ba%7D%20-%203%5Cmathbf%7Bb%7D%20-%202%5Cmathbf%7Bc%7D%20%3D%20%5Cmathbf%7B0%7D

, we obtain:

2%5Cmathbf%7Bc%7D%20%3D%205%5Cmathbf%7Ba%7D%20-%203%5Cmathbf%7Bb%7D.

Step 3: Substitute the Expression for $%5Cmathbf%7Bc%7D$
Substitute the previously derived expression for

%5Cmathbf%7Bc%7D

2%20%5Ctimes%20%5Cfrac%7B5%5Cmathbf%7Ba%7D%20-%203%5Cmathbf%7Bb%7D%7D%7B5%20-%203%7D%20%3D%205%5Cmathbf%7Ba%7D%20-%203%5Cmathbf%7Bb%7D.

Simplifying the expression gives:

%5Cmathbf%7Bc%7D%20%3D%20%5Cfrac%7B5%5Cmathbf%7Ba%7D%20-%203%5Cmathbf%7Bb%7D%7D%7B2%7D.

Thus, point

C

divides the line segment

BA

externally in the ratio

5%3A3

Step 1: Define the Points
Let the semicircle be centered at $O$ with radius $r$ , and points $A(-r%2C0)$ , $B(r%2C0)$ , and $P(x%2Cy)$ on the semicircle.
Step 2: Write the Vectors
Vectors from $A$ and $B$ to $P$ : $%5Cmathbf%7BAP%7D%20%3D%20(x%20%2B%20r%2C%20y)%2C%20%5Cquad%20%5Cmathbf%7BBP%7D%20%3D%20(x%20-%20r%2C%20y).$
Step 3: Find the Dot Product
$%5Cmathbf%7BAP%7D%20%5Ccdot%20%5Cmathbf%7BBP%7D%20%3D%20(x%20%2B%20r%2C%20y)%20%5Ccdot%20(x%20-%20r%2C%20y).$ $%3D%20(x%20%2B%20r)(x%20-%20r)%20%2B%20y%5E2.$ $%3D%20x%5E2%20-%20r%5E2%20%2B%20y%5E2.$ Using the equation of the semicircle: $x%5E2%20%2B%20y%5E2%20%3D%20r%5E2.$ $%5Cmathbf%7BAP%7D%20%5Ccdot%20%5Cmathbf%7BBP%7D%20%3D%20r%5E2%20-%20r%5E2%20%3D%200.$ Since dot product is zero, $AP$ is perpendicular to $BP$ , proving a right angle.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Vectors

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs