Geometrical applications MCQ Quiz in தமிழ் - Objective Question with Answer for Geometrical applications - இலவச PDF ஐப் பதிவிறக்கவும்

Given:

Three out of the four vertex of the parallelogram -

A = (1, 2, 3)

B = (-1, -2, -1)

C = (2, 3, 2)

Concept:

The diagonals of a paralleogram bisect each other.

Formula:

The mid-point of the line segment joining A (x1, y1, z1) and B (x2, y2, z2) is given by -

M (x, y, z) where \(x = \frac{x_1 + x_2}{2}\), \(y = \frac{y_1 + y_2}{2}\) and \(z = \frac{z_1 + z_2}{2}\)

Solution:

Let the midpoint of AC be M, then it is also the midpoint of BD

Now, M = (\(\frac{1+2}{2}\), \(\frac{2+3}{2}\), \(\frac{3+2}{2}\))

= (3/2, 5/2, 5/2)

Let D = (a, b, c)

Then, the midpoint of BD is given by (\(\frac{a-1}{2}\), \(\frac{b-2}{2}\), \(\frac{c-1}{2}\))

Equating it to M, we get a = 4, b = 7, c = 6

Hence D = (4, 7, 6)

Concept:

Triangle Law of Vector Addition: Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

\(\Rightarrow {\rm{\vec R}} = {\rm{\vec A}} + {\rm{\vec B}}\)

Calculation:

In triangle ABC,

\( \Rightarrow \overrightarrow {{\rm{BA}}} {\rm{\;}} + \overrightarrow {{\rm{AC}}} {\rm{\;}} = \overrightarrow {{\rm{BC}}} \)               …. (1)

In triangle BCD,

\(\Rightarrow \overrightarrow {{\rm{BC}}} + \overrightarrow {{\rm{CD}}} \; = \;\overrightarrow {{\rm{BD}}} \;\)            …. (2)

Put the value of BC vector in equation 2^nd,

\(\Rightarrow \overrightarrow {{\rm{BA}}} {\rm{\;}} + \overrightarrow {{\rm{AC}}} {\rm{\;}} + \overrightarrow {{\rm{CD}}} \; = \;\overrightarrow {{\rm{BD}}} \)

\(\Rightarrow \overrightarrow {{\rm{BA}}} - \overrightarrow {{\rm{CA}}} {\rm{\;}} + \overrightarrow {{\rm{CD}}} \; = \;\overrightarrow {{\rm{BD}}}\)                           \( \left(\because {\overrightarrow {{\rm{AC}}} = \; - \overrightarrow {{\rm{CA}}} } \right)\) 

\(\therefore \overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{BD}}} + \overrightarrow {{\rm{CA}}}\)

Calculation:

We have A(0,4,1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2)

∴ \(\overrightarrow{AB}=(2-0) \hat{i}+(3-4) \hat{j}+(-1-1) \hat{k}=2 \hat{i}-\hat{j}-2 \hat{k}\)

\(\overrightarrow{B C}=(4-2) \hat{i}+(5-3) \hat{j}+(0+1) \hat{k}=2 \hat{i}+2 \hat{j}+\hat{k}\)

\(\overrightarrow{C D}=(2-4) \hat{i}+(6-5) \hat{j}+(2-0) \hat{k}=-2 \hat{i}+\hat{j}+2 \hat{k}\)

\(\overrightarrow{D A}=(0-2) \hat{i}+(4-6) \hat{j}+(1-2) \hat{k}=-2 \hat{i}-2 \hat{j}-\hat{k}\)

∴ Area of quadrilateral ABCD:

= \(|\overrightarrow{A B} \times \overrightarrow{B C}|\)

= \(\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{array}\right|\)

= \(|\hat i(-1+4)-\hat j(2+4)+\hat k(4+2)|\)

= \(|3 \hat{i}-6 \hat{j}+6 \hat{k}|\)

= \(\sqrt{9+36+36}\)

= 9 sq. units

∴ The area of the quadrilateral ABCD is equal to 9 sq. units.

The correct answer is option 1

Concept:

Calculations:

Given, triangle whose vertices are at A = (3, -1, 2) = \(\rm 3\vec i -\vec j + 2\vec k\)

B = (1, -1, -3) =  \(\rm \vec i -\vec j - 3\vec k\)

and C = (4, -3, 1) =  \(\rm 4\vec i -3\vec j + \vec k\)

Let A, B , and C be the vertices of the \(\triangle ABC\) , then the area of that triangle = \(\rm \dfrac 1 2 \times |\vec {AB}\times \vec {AC}|\)

\(\rm \vec {AB} = \rm (\vec i -\vec j - 3\vec k) - (3\vec i -\vec j + 2\vec k)\)

⇒\(\rm \vec {AB}= -2\vec i + 0\vec k -5 \vec k\)

\(\rm \vec {AC} = \rm (4\vec i -3\vec j + \vec k) - (3\vec i -\vec j + 2\vec k)\)

⇒ \(\rm \vec {AC} = \vec i -2\vec j -\vec k\)

Now, \(\rm \vec {AB}\times\vec {AC}= \)\(\begin{vmatrix} \vec i&\vec j & \vec k \\ -2& 0 & -5 \\ 1 & -2 & -1 \end{vmatrix}\)

⇒\(\rm \vec {AB} \times \vec {AC} = \vec i (-10)-\vec j(2+5)+\vec k(4-0)\)

⇒\(\rm \vec {AB} \times \vec {AC} = -10\vec i -7\vec j+4\vec k\)

⇒ \(\rm |\vec {AB} \times \vec {AC}| = \sqrt {(-10)^2+(-7)^2+(4)^2 }\)

⇒\(\rm |\vec {AB} \times \vec {AC}| = \sqrt {165}\)

The area of that triangle = \(\dfrac 1 2 \times |\vec {AB}\times \vec {AC}|\)

⇒ The area of that triangle = \(\dfrac 12 \times \sqrt {165}\)
Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is \(\dfrac {\sqrt {165}}{2}\)

Given

x = t² + 2, y = 4t - 5, z = 2t² - 6t

unit tangent vector is given as,

\(\vec r\frac{{\frac{{dx}}{{dt}}\hat i + \frac{{dy}}{{dt}}\hat j + \frac{{dz}}{{dt}}\hat k}}{{\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left(r {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}}}}\)

\({\left. {\frac{{dx}}{{dt}}} \right|_{t = 2}} = \frac{d}{{dt}}({t^2} + 2) = 2t = 4\)

\({\left. {\frac{{dy}}{{dt}}} \right|_{t = 2}} = \frac{d}{{dt}}(4t - 5) = 4\)

\({\left. {\frac{{dz}}{{dt}}} \right|_{t = 2}} = \frac{d}{{dt}}(2{t^2} - 6t)\)  = 4t - 6 = 4 × 2 - 6 = 2

\(\vec r = \frac{{4\hat i + 4\hat j + 2\hat k}}{{\sqrt {{4^2} + {4^2} + {2^2}} }}\)\( = \frac{{4\hat i + 4\hat j + 2\hat k}}{{\sqrt {36} }}\)

\(\vec r = \frac{{2\left( {2\hat i + 2\hat j + \hat k} \right)}}{6}\)

\(\vec r = \frac{1}{3} \left( {2\hat i + 2\hat j + \hat k} \right)\)

Concept:

If the position vectors of points A and B are \(\rm \vec {a}\) and \(\rm \vec {b}\) respectively, then \(\rm \vec {AB}=\vec b - \vec a\).

Area of a Triangle with vectors \(\rm \vec {AB}\) and \(\rm \vec {AC}\) as its sides is given by: \(\rm Area=\dfrac12|\vec{A}\times\vec{B}|\).

Cross Product: For two vectors \(\rm \vec {A}=a_1\hat i+a_2\hat j+a_3\hat k\) and \(\rm \vec {B}=b_1\hat i+b_2\hat j+b_3\hat k\), their cross product is given by:

• \(\rm \vec A \times \vec B=\begin{vmatrix} \rm \hat i & \rm \hat j & \rm \hat k\\ \rm a_1& \rm a_2 & \rm a_3\\ \rm b_1 & \rm b_2 & \rm b_3\end{vmatrix}=(a_2b_3-a_3b_2)\hat i+(a_3b_1-a_1b_3)\hat j+(a_1b_2-a_2b_1)\hat k\).
• \(\rm \vec A\times\vec B=-\left(\vec B \times \vec A\right)\).
• \(\rm \vec A\times\vec A=0\).

The magnitude \(\rm |\vec A|\) of a vector \(\rm \vec {A}=a_1\hat i+a_2\hat j+a_3\hat k\) is given by: \(\rm |\vec A|=\sqrt{a_1^2+a_2^2+a_3^2}\).

Calculation:

We have \(\rm \vec {AB}=\vec b - \vec a\) and \(\rm \vec {AC}=\vec a - \vec c\).

Using the formula for the area of a triangle:

\(\rm Area=\dfrac{1}{2}\left|\vec {AB}\times\vec {AC}\right|\)

= \(\rm \dfrac{1}{2}\left|\left(\vec b-\vec a\right)\times\left(\vec c-\vec a\right)\right|\)

= \(\rm \dfrac{1}{2}\left|\vec b\times\vec c-\vec b\times\vec a -\vec a\times\vec c+\vec a\times\vec a\right|\)

= \(\rm \dfrac{1}{2} |\vec{a}\times \vec{b} + \vec{b} \times \vec{c}+\vec{c}\times \vec{a}|\).

Additional Information

Area of a parallelogram with vectors \(\rm \vec {a}\) and \(\rm \vec {b}\) as its sides is given by: \(\rm Area=\left|\vec{a}\times\vec{b}\right|\).

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

• Dot Product is defined as: \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
• Cross Product is defined as: \(\rm \vec A\times \vec B=\hat n|\vec A||\vec B|\sin \theta\), where \(\rm \hat n\) is the unit vector perpendicular to the plane containing \(\rm \vec A\) and \(\rm \vec B\).

Concept:

\({\rm{\vec a}}.{\rm{\vec b}}= |{\rm{\vec a}}|.|{\rm{\vec b}}|\cos \theta\)

Calculation:

Let, \(\vec{a}=\frac{1}{{\sqrt {2} }}{\rm{\hat i}} - \frac{1}{{\sqrt 2 }}{\rm{\hat j}} + {\rm{\hat k}}\) and \(\vec{b}=\frac{1}{{\sqrt {2} }}{\rm{\hat i}} + \frac{1}{{\sqrt 2 }}{\rm{\hat j}} + {\rm{\hat k}}\)

\(\begin{array}{l} \operatorname{cos\theta}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\ \\ {\vec{a} \cdot \vec{b}}=\left(\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}+\hat{k}\right) \cdot\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\hat{k}\right) \\ = \frac{1}{2}-\frac{1}{2}+1 \\ =1 \end{array}\)

And,

\(\begin{aligned} |\vec{a}| &=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(-\frac{1}{\sqrt{2}}\right)^{2}+(1)^{2}} \\ &=\sqrt{\frac{1}{2}+\frac{1}{2}+1} \\ &=\sqrt{2} \end{aligned}\\ \begin{aligned} |\vec{b}| &=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+(1)^{2}} \\ &=\sqrt{\frac{1}{2}+\frac{1}{2}+1} \\ &=\sqrt{2} \end{aligned}\)

\(\begin{array}{l} \therefore\cos \theta=\frac{1}{\sqrt{2} \sqrt{2}}=\frac{1}{2} \\ \Rightarrow\theta=60^{\circ} \end{array}\)

Hence, option (1) is correct.

Concept:

\(\begin{array}{l} \vec{a}=x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k}\\ \vec{b}=x_{2} \hat{i}+y_{2} \hat{j}+z_{2} \hat{k} \\ \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & j & \hat{k} \\ x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \end{array}\right| \end{array}\)

Calculation:

We have, \(\overrightarrow {{\rm{OA}}} = 3{\rm{\;\hat i}} - {\rm{\;\hat j}} + {\rm{\;\hat k}}\)   and \(\overrightarrow {{\rm{OB}}} = 2{\rm{\hat i}} + {\rm{\;\hat j}} - 3{\rm{\hat k}}\)

\(\begin{aligned} \overrightarrow{O A} \times \overrightarrow{O B} &=\left|\begin{array}{ccc} \hat{i} & \hat{j}&\hat{k}\\ 3 & -1 & 1 \\ 2 & 1 & -3 \end{array}\right| \\ &=i(3-1)-\hat{j}(-9-2)+\widehat{k}(3+2) \\ &=2 \hat{i}+11 \hat{j}+5 \hat{k} \end{aligned}\)

Now, 

\(\begin{aligned} |\overrightarrow{OA} \times \overrightarrow{O B}| &=\sqrt{(2)^{2}+(11)^{2}+(5)^{2}} \\ &=\sqrt{4+121+25} \\ &=\sqrt{150} \\&=5√6 \end{aligned}\)

Area of triangle =  \(\frac{1}{2}|\overrightarrow{OA} \times \overrightarrow{O B}|\)

= 1/2(5√6)

= ​​\(\frac{{5\sqrt 6 }}{2} \)square unit

Hence, option (2) is correct.

Calculation:

By vector law,              

Similarly,    \(\begin{array}{l} \overrightarrow{O A}+\overrightarrow{A P}=\overrightarrow{O P} \\ \overrightarrow{O B}+\overrightarrow{B P}=\overrightarrow{O P} \\ \overrightarrow{O C}+\overrightarrow{C P}=\overrightarrow{O P} \\ \overrightarrow{O D}+\overrightarrow{D P}=\overrightarrow{O P} \end{array}\)

Now, adding the above equations, we get 

\(\begin{array}{l} \overrightarrow{O A}+\overrightarrow{A P}+\overrightarrow{O B}+\overrightarrow{B P}+\overrightarrow{O C}+\overrightarrow{C P}+\overrightarrow{O D}+\overrightarrow{D P}=4 \overrightarrow{O P} \\ \Rightarrow(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+\overrightarrow{O D})+(\overrightarrow{A P}+\overrightarrow{B P}+\overrightarrow{C P}+\overrightarrow{D P})=4 \overrightarrow{O P} \end{array}\)

From fig, AP = - BP and CP = -DP

∴ \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} \) 

= 4 OP 

 Hence, option (1) is correct.

Concept:

Area of rectangle ABCD = |AB| × |BC|

{Product of adjacent sides}

Calculation:

A = \(\rm -\hat{i} + \frac{1}{2} \hat {j} + 4 \hat {k},\) B = \(\rm \hat{i} + \frac{1}{2} \hat {j} + 4 \hat {k} \)

⇒ AB = \(\rm (1-(-1))\hat{i} +( \frac{1}{2} -\frac{1}{2})\hat {j} + (4-4) \hat {k}, \)

⇒ AB = \(2\hat {i}\)

⇒ |AB| = 2

Similarly,

B = \(\rm \hat{i} +\frac{1}{2} \hat {j} + 4 \hat {k} \), C = \(\rm \hat{i} - \frac{1}{2} \hat {j} + 4 \hat {k},\)

⇒ BC = \(\rm (1-1)\hat{i} +( -\frac{1}{2} -(\frac{1}{2}))\hat {j} + (4-4) \hat {k}, \)

⇒ BC = \(-\hat {j}\)

⇒ |BC| = 1

⇒ Area of rectangle ABCD = |AB| × |BC|

⇒ Area of rectangle ABCD = 2 × 1 = 2 sq unit

∴ The correct answer is option (3).