An instructor at the astronomical centre shows three among

Q 1:- An instructor at the astronomical centre shows three among the brightest stars in a particular constellation. Assume that the telescope is located at O(0, 0, 0) and the three stars have their locations at the points D, A and V having position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}, 7\hat{i} + 5\hat{j} + 8\hat{k}$ and $-3\hat{i} + 7\hat{j} + 11\hat{k}$ respectively. [CBSE 2024]

Based on the above information, answer the following question:

(i) How far is the star V from star A ? (1)

(ii) Find a unit vector in the direction of $\vec{DA}$ . (1)

(iii)(a) Find the measure of ∠VDA . (2)

(iii) (b) What is the projection of vector $\vec{DV}$ on vector $\vec{DA}$ ? (2)

Solution :- $\vec{OD} =2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{OA} = 7\hat{i} + 5\hat{j} + 8\hat{k}$

$\vec{OV} = -3\hat{i} + 7\hat{j} + 11\hat{k}$

(i) $\vec{VA} = \vec{OA} - \vec{OV}$

⇒ $\vec{VA} = ( 7\hat{i} + 5\hat{j} + 8\hat{k}) - (-3\hat{i} + 7\hat{j} + 11\hat{k})$

⇒ $\vec{VA} = 7\hat{i} + 5\hat{j} + 8\hat{k} + 3\hat{i} - 7\hat{j} - 11\hat{k}$

⇒ $\vec{VA} = 10\hat{i} - 2\hat{j} - 3\hat{k}$

Now $|\vec{VA}| = \sqrt{(10)^2 + (-2)^2 + (-3)^2}$

⇒ $|\vec{VA}| = \sqrt{100 + 4 + 9} = \sqrt{113}$

⇒ $|\vec{VA}| = \sqrt{113}$

Thus, distance between V to A = √113 unit

(ii) Since, $\vec{OD} =2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{OA} = 7\hat{i} + 5\hat{j} + 8\hat{k}$

$\vec{DA} = \vec{OA} - \vec{OD}$

⇒ $\vec{DA} = ( 7\hat{i} + 5\hat{j} + 8\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})$

⇒ $\vec{DA} = 7\hat{i} + 5\hat{j} + 8\hat{k} - 2\hat{i} - 3\hat{j} - 4\hat{k}$

⇒ $\vec{DA} = 5\hat{i} + 2\hat{j} + 4\hat{k}$

Now, $|\vec{DA}| = \sqrt{(5)^2 + (2)^2 + (4)^2}$

⇒ $|\vec{DA}| = \sqrt{25 + 4 + 16} = \sqrt{45}$

⇒ $|\vec{DA}| = 3\sqrt{5}$

Now, $\hat{DA} = \dfrac{\vec{DA}}{|\vec{DA}|}$

$= \dfrac{5\hat{i} + 2\hat{j} + 4\hat{k}}{ 3\sqrt{5}}$

(iii)(a) Since, $\vec{OD} =2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{OA} = 7\hat{i} + 5\hat{j} + 8\hat{k}$

$\vec{OV} = -3\hat{i} + 7\hat{j} + 11\hat{k}$

$\vec{DV} = \vec{OV} - \vec{OD}$

⇒ $\vec{DV} = ( -3\hat{i} + 7\hat{j} + 11\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})$

⇒ $\vec{DV} = -3\hat{i} + 7\hat{j} + 11\hat{k} - 2\hat{i} - 3\hat{j} - 4\hat{k}$

⇒ $\vec{DV} = -5\hat{i} + 4\hat{j} + 7\hat{k}$

$|\vec{DV}| = \sqrt{(-5)^2 + (4)^2 + (7)^2}$

⇒ $|\vec{DV}| = \sqrt{25 + 16 + 49} = \sqrt{45}$

⇒ $|\vec{DV}| = \sqrt{90} = 3\sqrt{10}$

Similarly,

$\vec{DA} = \vec{OA} - \vec{OD}$

⇒ $\vec{DA} = ( 7\hat{i} + 5\hat{j} + 8\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})$

⇒ $\vec{DA} = 7\hat{i} + 5\hat{j} + 8\hat{k} - 2\hat{i} - 3\hat{j} - 4\hat{k}$

⇒ $\vec{DA} = 5\hat{i} + 2\hat{j} + 4\hat{k}$

Now, $|\vec{DA}| = \sqrt{(5)^2 + (2)^2 + (4)^2}$

⇒ $|\vec{DA}| = \sqrt{25 + 4 + 16} = \sqrt{45}$

⇒ $|\vec{DA}| = 3\sqrt{5}$

$\vec{DA}.\vec{DV} = (5\hat{i} + 2\hat{j} + 4\hat{k}).(-5\hat{i} + 4\hat{j} + 7\hat{k})$

= -25 + 8 + 28 = 11

Now, $\cos (\angle VDA) = \dfrac{\vec{DA}.\vec{DV}}{|\vec{DA}||\vec{DV}|}$

$= \dfrac{11}{3\sqrt{5}.3\sqrt{10}} = \dfrac{11}{9\sqrt{50}}$

$\cos (\angle VDA) = \dfrac{11}{18\sqrt{2}}$

⇒ $\angle VDA = \cos^{-1} (\dfrac{11}{18\sqrt{2}})$

(iii)(b) The projection of vector $\vec{DV}$ on vector $\vec{DA}$

$= \dfrac{\vec{DV}.\vec{DA}}{|\vec{DA}|}$

$= \dfrac{(5\hat{i} + 2\hat{j} + 4\hat{k}).(-5\hat{i} + 4\hat{j} + 7\hat{k})}{3\sqrt{5}}$

$= \dfrac{11}{3\sqrt{5}}$

Hence, The projection of vector $\vec{DV}$ on vector $\vec{DA} = \dfrac{11}{3\sqrt{5}}$

Q 2: Rohit Jaspreet and Alia appeared for an interview for three vacancies in the same post. The probability of Rohit’s selection is 1/5, Jaspreet’s selection is 1/3 and Alia’s selection is 1/4. The event of selection is independent of each other. [CBSE 2024]

Based on the above information, answer the following question :

(i) What is the probability that at least one of them is selected ? (1)

(ii) Find $P(G/\overline{H})$ where G is the event of Jaspreet’s selection and $\bar{H}$ denotes the event that Rohit is not selected. (1)

(iii) Find the probability that exactly one of them is selected. (2)

(iii) (b) Find the probability that exactly two of them are selected. (2)

Solution:- See full solution

Q 3:- A store has been selling calculators at Rs 350 each. A market survey indicates that a reduction in price (p) of calculator increases the number of units (x) sold. The relation between the price and quantity sold is given by the demand function $p = 450 - \frac{1}{2} x$ . [CBSE 2024]

Based on the above information, answer the following questions:

(i) Determine the number of units (x) that should be sold to maximise the revenue R(x) = xp(x). Also, verify result. (2)

(ii) What rebate in price of calculator should the store give to maximise the revenue ? (2)

Solution:- See full solution

Q 1:- In a group activity class, there are 10 students whose ages are 16, 17, 15, 14, 19, 17, 16, 19, 16 and 15 years. One student is selected at random such that each has equal chance of being choosen and age of the student is recorded. [CBSE 2023]

On the basis of the above information, answer the following question :

(i) Find the probability that the age of the selected student is a composite number. (1)

(ii) Let X be the age of the selected student. What can be the value of X ? (1)

(iii) (a) Find the probability distribution of random variable X and hence find the mean age. (2)

(iii) (b) A student was selected at random and his age was to be greater than 15 years. Find the probability that his age is a prime number. (2)

Solution:- See full solution

Q:- A building contractor undertakes a job to construct 4 flats on a plot along with the parking area. Due to strike the probability of many construction workers not being present present for the job is 0.65. The probability that many are not present and still the work get completed on times is 0.35. The probability that work will be completed on time when all workers are present is 0.80. [CBSE 2023]

Solution:- See full solution

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