SATHEE: 3D Geometry 1 Question 3

3D Geometry 1 Question 3

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: (a)

Solution:

Method 1:

Let’s approach this step-by-step:

First, we need to find the coordinates of points B and C. We know they lie on the line given by the equation:

$\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$
This line passes through the point (-2, 1, 0) and is parallel to the vector (3, 0, 4). So we can parameterize the line as:

$B (- 2 + 3 t, 1, 4 t)$ and $C (- 2 + 3 s, 1, 4 s)$ , where $t$ and $s$ are real numbers.
Now, we’re told that the distance between B and C is 5 units. Using the distance formula in 3D, we get:

$\sqrt{(3 s - 3 t)^{2} + (4 s - 4 t)^{2}} = 5$

$\sqrt{9 (s - t)^{2} + 16 (s - t)^{2}} = 5$

$5 | s - t | = 5$

$| s - t | = 1$
Without loss of generality, let’s choose $t = 0$ and $s = 1$ . Then the coordinates of B and C are:

$B (- 2, 1, 0)$ and $C (1, 1, 4)$
Now we have the coordinates of all three vertices of the triangle:

$A (1, - 1, 2)$ , $B (- 2, 1, 0)$ , and $C (1, 1, 4)$
To find the area of the triangle, we can use the formula:

$A r e a = \frac{1}{2} | | (B - A) \times (C - A) | |$

where $\times$ denotes the cross product and $| |$ denotes the magnitude of a vector.
$B - A = (- 3, 2, - 2)$ and $C - A = (0, 2, 2)$

$(B - A) \times (C - A) = (- 8, - 6, 6)$

$| | (B - A) \times (C - A) | | = \sqrt{(- 8)^{2} + (- 6)^{2} + 6^{2}} = \sqrt{136} = 2 \sqrt{34}$
Therefore, the area of the triangle is:

$A r e a = \frac{1}{2} \cdot 2 \sqrt{34} = \sqrt{34}$ square units.

So the correct answer is (a) $\sqrt{34}$ square units[1].

Alternate Method

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 $△ A B C$

$= \frac{1}{2} | B C | | b | | \sin θ |$

[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)

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

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

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3D Geometry 1 Question 3

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

Answer:

Alternate Method

Alternate Method

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3D Geometry 1 Question 3

The vertices B and C of a △ABC lie on the line, x+23=y−10=z4 such that BC=5 units. Then, the area (in sq units) of this triangle, given that the point A(1,−1,2) is

Answer:

Alternate Method

Alternate Method

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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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