Denote the origin by O, (2,1) by P and (–1,2) by Q. If R and S are marked on the XY-plane such that OR = 30 units, OS = 40 units, P lies on OR and Q on OS, then what is the distance between R and S, in the given unit?
Show Hint
- 50
- 70
- 57
- 64
The Correct Option is A
Solution and Explanation
We are given the coordinates of the origin O, point P, and point Q. Point R lies on the line segment OR, which also contains P. Point S lies on the line segment OS, which also contains Q. We are given the distances OR and OS. We need to find the distance between R and S.
Step 2: Key Formula or Approach:
The problem can be visualized as a triangle ORS with two sides OR and OS given. If we can find the angle between these two sides, \(\angle ROS\), we can use the Law of Cosines to find the third side RS. The angle \(\angle ROS\) is the same as the angle \(\angle POQ\). We can find the angle between the vectors \(\vec{OP}\) and \(\vec{OQ}\) using the dot product formula.
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \] \[ RS^2 = OR^2 + OS^2 - 2(OR)(OS)\cos(\angle ROS) \] Step 3: Detailed Explanation:
The position vectors for P and Q are \(\vec{OP} = (2, 1)\) and \(\vec{OQ} = (-1, 2)\).
Let's find the dot product of these two vectors to find the angle between them.
\[ \vec{OP} \cdot \vec{OQ} = (2)(-1) + (1)(2) = -2 + 2 = 0 \] Since the dot product is 0, the vectors \(\vec{OP}\) and \(\vec{OQ}\) are perpendicular to each other.
This means the angle between the lines OR and OS, \(\angle ROS\), is 90\(^{\circ}\).
Therefore, \(\triangle ORS\) is a right-angled triangle with the right angle at the origin O.
The sides OR and OS are the two legs of the right triangle, and the distance RS is the hypotenuse.
Using the Pythagorean theorem:
\[ RS^2 = OR^2 + OS^2 \] We are given OR = 30 and OS = 40.
\[ RS^2 = 30^2 + 40^2 = 900 + 1600 = 2500 \] \[ RS = \sqrt{2500} = 50 \] Step 4: Final Answer:
The distance between R and S is 50 units.
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Questions Asked in SRCC GBO exam
Read the given passage and answer the questions.
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