Question
Prove by vector method, that the angle subtended on semicircle is a right angle.
✓
Answer
Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B.
Then $\angle A P B$ is an angle subtended on a semicircle.
Let $\overline{ AC }=\overline{ CB }=\overline{ a }$ and $\overline{ CP }=\overline{ r }$
Then $|\overline{ a }|=|\overline{ r }|$ ....(1)
$\overline{ AP }=\overline{ AC }+\overline{ CP }$
$=\overline{ a }+\overline{ r }$
$=\bar{r}+\bar{a}$
$\overline{ BP }=\overline{ BC }+\overline{ CP }$
$=-\overline{ CB }+\overline{ CP }$
$=-\bar{a}+\bar{r}$
$\therefore \overline{ AP } \cdot \overline{ BP }=(\overline{ r }+\overline{ a }) \cdot(\overline{ r }-\overline{ a })$
$=\overline{ r } \cdot \overline{ r }-\overline{ r } \cdot \overline{ a }+\overline{ a } \cdot \overline{ r }-\overline{ a } \cdot \overline{ a }$
$=|\overline{ r }|^2-|\overline{ a }|^2$
$=0 \quad \ldots(\because \bar{r} \cdot \bar{a}=\bar{a} \cdot \bar{r})$
$\therefore \overline{ AP } \perp \overline{ BP }$
$\therefore \angle APB$ is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
Then $\angle A P B$ is an angle subtended on a semicircle.
Let $\overline{ AC }=\overline{ CB }=\overline{ a }$ and $\overline{ CP }=\overline{ r }$
Then $|\overline{ a }|=|\overline{ r }|$ ....(1)
$\overline{ AP }=\overline{ AC }+\overline{ CP }$
$=\overline{ a }+\overline{ r }$
$=\bar{r}+\bar{a}$
$\overline{ BP }=\overline{ BC }+\overline{ CP }$
$=-\overline{ CB }+\overline{ CP }$
$=-\bar{a}+\bar{r}$
$\therefore \overline{ AP } \cdot \overline{ BP }=(\overline{ r }+\overline{ a }) \cdot(\overline{ r }-\overline{ a })$
$=\overline{ r } \cdot \overline{ r }-\overline{ r } \cdot \overline{ a }+\overline{ a } \cdot \overline{ r }-\overline{ a } \cdot \overline{ a }$
$=|\overline{ r }|^2-|\overline{ a }|^2$
$=0 \quad \ldots(\because \bar{r} \cdot \bar{a}=\bar{a} \cdot \bar{r})$
$\therefore \overline{ AP } \perp \overline{ BP }$
$\therefore \angle APB$ is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following equation for x:
$\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{2\pi}{3},\text{x}>0$
→$\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{2\pi}{3},\text{x}>0$
If the mean of a binomial distribution is 20 and its standard deviation is 4, find p.
Of the students in a college, it is known that $60 \%$ reside in a hostel and $40 \%$ do not reside in hostel. Previous year results report that $30\%$ of students residing in hostel attain A grade and $20\%$ of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?
→Find the general solution of the following differential equations: $\frac{d y}{d x}=x \sqrt{25-x^2}$
→Using the truth table, prove the following logical equivalence : $p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
→Diffrentiate the following w.r.t.x
→$\left(x^2+4 x+1\right)^3+\left(x^3-5 x-2\right)^4$
Using determinants, find the equation of the line joining the points:
$(3, 1)$ and $(9, 3)$
→$(3, 1)$ and $(9, 3)$
Define $\vec{\text{a}}\times\vec{\text{b}}$ and prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta,$ where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
→Evaluate $\sin\Big(\frac{1}{2}\sin^{-1}\frac{4}{5}\Big).$
→Find the co-ordinates of the point which divides the line segment joining the points $\mathrm{A}(2,-6,8)$ and $\mathrm{B}(-1,3,-4)$. (i) Internally in the ratio $1: 3$. (ii) Externally in the ratio $1: 3$.
→