SATHEE: 3D Geometry 2 Question 2

3D Geometry 2 Question 2

####2. The vertices $B$ and $C$ of a $△ A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then, the area (in sq units) of this triangle, given that the point $A (1, - 1, 2)$ is

(2019 Main, 9 April II)

(a) $\sqrt{34}$

(b) $2 \sqrt{34}$

(d) 6

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Answer:

Correct Answer: 2. (a)

Solution:

Given line is $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$

Vector along line is, $a = 3 \hat{i} + 4 \hat{k}$

and vector joining the points $(1, - 1, 2)$ to $(- 2, 1, 0)$ is $b = (1 + 2) \hat{i} + (- 1 - 1) \hat{j} + (2 - 0) \hat{k}$

$= 3 \hat{j} - 2 \hat{j} + 2 \hat{k}$

and $| B C | = 5$ units

Now, area of required $△ A B C$

$= \frac{1}{2} | B C | | b | | \sin θ |$ …… (i)

[where $θ$ is angle between vectors $a$ and $b$ ]

$∵ | b | \sin θ = \frac{| a \times b |}{| a |}$ ,

$∵ | a \times b | = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 4 \\ 3 & - 2 & 2 \end{array} | = 8 \hat{i} + 6 \hat{j} - 6 \hat{k}$

$\Rightarrow | a \times b | = \sqrt{64 + 36 + 36}$

$= \sqrt{136} = 2 \sqrt{34}$

and $| a | = \sqrt{9 + 16} = 5$

$∴ | b | \sin θ = \frac{2 \sqrt{34}}{5}$

On substituting these values in Eq. (i), we get

Required area $= \frac{1}{2} \times 5 \times \frac{2 \sqrt{34}}{5} = \sqrt{34}$ sq units

Alternate Method

Given line is $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4} = λ$ (let) …(i)

Since, point $D$ lies on the line $B C$ .

$∴$ Coordinates of $D = (3 λ - 2, 1, 4 λ)$

Now, $D R$ of $B C \Rightarrow a_{1} = 3, b_{1} = 0, c_{1} = 4$

and $D R$ of $A D \Rightarrow a_{2} = 3 λ - 3, b_{2} = 2, c_{2} = 4 λ - 2$

Since, $A D ⊥ B C, a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$3 \times (3 λ - 3) + 0 (2) + 4 (4 λ - 2) = 0$

$\Rightarrow 9 λ - 9 + 0 + 16 λ - 8 = 0$

$\Rightarrow 25 λ - 17 = 0$

$\Rightarrow λ = \frac{17}{25}$

$∴$ Coordinates of $D = \frac{1}{25}, 1, \frac{68}{25}$ .

Now, $A D = \sqrt{1 - {\frac{1}{25}}^{2} + (- 1 - 1)^{2} + 2 - {\frac{68}{25}}^{2}}$

$= \sqrt{{\frac{24}{25}}^{2} + (- 2)^{2} + {\frac{- 18}{25}}^{2}}$

$= \sqrt{\frac{576}{625} + 4 + \frac{324}{625}} = \frac{2}{5} \sqrt{34}$

$∴$ Area of $△ A B C = \frac{1}{2} B C \times A D$

$\begin{aligned} = \frac{1}{2} \times 5 \times \frac{2}{5} \sqrt{34} \\ = \sqrt{34} sq units \end{aligned}$

3D Geometry 2 Question 2

Answer:

Alternate Method

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3D Geometry 2 Question 2

Answer:

Alternate Method

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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