Geometrical applications MCQ Quiz - Objective Question with Answer for Geometrical applications - Download Free PDF

Calculation:

Given,

Points A(k, 1, -1), B(2k, 0, 2), and C(2 + 2k, k, 1) are the vertices of the triangle.

The vectors AB and BC are given by:

\( \overrightarrow{AB} = B - A = (2k - k, 0 - 1, 2 - (-1)) = (k, -1, 3) \)

\( \overrightarrow{BC} = C - B = (2 + 2k - 2k, k - 0, 1 - 2) = (2, k, -1) \)

The dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) is:

\( \overrightarrow{AB} \cdot \overrightarrow{BC} = (k)(2) + (-1)(k) + (3)(-1) = 2k - k - 3 = k - 3 \)

Since the vectors are perpendicular, their dot product must be zero:

\( k - 3 = 0 \)

Thus, \( k = 3 \).

∴ The value of k is \(3\).

Hence, the correct answer is Option 4. 

Position vector of centriod \(\vec {G} = \dfrac {\vec {a} + \vec {b} + \vec {c}}{3}\)

Position vector of circum center \(\vec {C} = \dfrac {\vec {a} + \vec {b} + \vec {c}}{4}\)

Apply Section Formula,

\(\vec {G} = \dfrac {2\vec {C} + \vec {R}}{3}\)

\(3\vec {G} = 2\vec {C} + \vec {R}\)

\(\vec {R} = 3\vec {G} - 2\vec {C} = (\vec {a} + \vec {b} + \vec {c}) - 2 \left (\dfrac {\vec {a} + \vec {b} + \vec {c}}{4}\right )\)

\(= \dfrac {\vec {a} + \vec {b} + \vec {c}}{2}\)

Calculation:

Let the position vector of A, B, C, D, E, F be a̅ , b̅, c̅, d̅ , e̅, f̅ respectively.

∴ \(\overline{\mathrm{d}}=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}, \overline{\mathrm{e}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}, \overline{\mathrm{f}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}\)

Now, \(\overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}}\)

= \(\overline{\mathrm{d}}-\overline{\mathrm{a}}+\frac{2}{3}(\overline{\mathrm{e}}-\overline{\mathrm{b}})+\frac{1}{3}(\overline{\mathrm{f}}-\overline{\mathrm{c}})\)

= \(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}-\overline{\mathrm{a}}+\frac{2}{3}\left(\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}-\overline{\mathrm{b}}\right)+\frac{1}{3}\left(\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}-\overline{\mathrm{c}}\right)\)

= \(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}-2 \overline{\mathrm{a}}}{2}+\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}}{3}+\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}-2 \overline{\mathrm{c}}}{6}\)

= \(\frac{3 \overline{\mathrm{c}}-3 \overline{\mathrm{a}}}{6}\)

= \(\frac{3}{6}(\overline{\mathrm{c}}-\overline{\mathrm{a}})\)

= \(\frac{1}{2} \overline{\mathrm{AC}}\)

∴ The required answer is \(\frac{1}{2} \overline{\mathrm{AC}}\).

The correct answer is Option 2.

Calculation 

Centroid of triangle ABC = \((\frac{1+3+2}{3} , \frac{3+1+4}{3}) = (2,\frac{8}{3})\)

Image of point \( (2,\frac{8}{3})\) in the line 𝑥 + 2𝑦 = 4

\(\frac{h-2}{1} = \frac{k-\frac{8}{3}}{2} = \frac{-2(1(2)+2(\frac{8}{3})-4)}{(1^{2}+2^{2})}\)

\(\frac{h-2}{1} = \frac{3k-8}{6} = \frac{-2(10)}{15}\)

\(h = 2 - \frac{4}{3} = \frac{2}{3}\)

\(3k = 8 - \frac{4}{3}\times 6 = 0\)

The centroid of the triangle DEF is (2/3, 0)

Hence option 4 is correct

Calculation:

Since ABC is an equilateral triangle, all angles are 60°.

Let \(\vec{AB}\) be along the x-axis and \(\vec{AC}\) at 60° from it.

Then \(\vec{AB} = a\hat{i}\) and \(\vec{AC} = a(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = \frac{a}{2}\hat{i} + \frac{a\sqrt{3}}{2}\hat{j}\)

We have \(\vec{AM} = \frac{1}{3}\vec{AB} = \frac{a}{3}\hat{i}\)

and \(\vec{AN} = K\vec{AC} = \frac{Ka}{2}\hat{i} + \frac{Ka\sqrt{3}}{2}\hat{j}\)

Now, \(\vec{BN} = \vec{AN} - \vec{AB} = (\frac{Ka}{2} - a)\hat{i} + \frac{Ka\sqrt{3}}{2}\hat{j}\)

and \(\vec{CM} = \vec{AM} - \vec{AC} = (\frac{a}{3} - \frac{a}{2})\hat{i} - \frac{a\sqrt{3}}{2}\hat{j} = -\frac{a}{6}\hat{i} - \frac{a\sqrt{3}}{2}\hat{j}\)

Since \(\vec{BN}\) and \(\vec{CM}\) are perpendicular, their dot product is zero.

\(\vec{BN} \cdot \vec{CM} = 0\)

⇒ \((\frac{Ka}{2} - a)(-\frac{a}{6}) + (\frac{Ka\sqrt{3}}{2})(-\frac{a\sqrt{3}}{2}) = 0\)

⇒ \(-\frac{Ka^2}{12} + \frac{a^2}{6} - \frac{3Ka^2}{4} = 0\)

⇒ \(-Ka + 2a - 9Ka = 0\)

⇒ \(10K = 2\)

⇒ \(K = \frac{1}{5}\)

Hence option 1 is correct

Concept:

If \(\rm \vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\rm \vec a \times \;\vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{ \rm a_1}}&{{\rm a_2}}&{{\rm a_3}}\\ {{\rm b_1}}&{{\rm b_2}}&{{\rm b_3}} \end{array}} \right|\)

If \(\rm \vec a\;and\;\vec b\) are the adjacent sides of a triangle, then the area of the triangle is given by: \(\rm \frac{1}{2}\;\left| {\vec a \times \;\vec b} \right|\)

Calculation:

Given: two sides of the triangle are  \(\left( {2\bar i - 7\bar j + \bar k} \right)\;\)and \(\left( {4\bar j - 3\bar k} \right)\)

To Find: Area of the triangle

Let sides be \(\rm \vec a\;and\;\vec b\)=  \(\left( {2\bar i - 7\bar j + \bar k} \right)\;\)and \(\left( {4\bar j - 3\bar k} \right)\)

\(\rm \vec{a}\times \vec{b}= \begin{vmatrix} {\hat i}&{\hat j}&{\hat k} \\ 2 & -7 & 1\\ 0 & 4 & -3 \end{vmatrix}\\=\hat i(21-4)-\hat j(-6-0)+\hat k(8-0)\\=17\hat i+6\hat j+8\hat k\)

\(\rm |\vec{a}\times \vec{b}|= \sqrt{17^2+6^2+8^2}= \sqrt{389}\)

Now,

Area of the triangle = \(\rm \frac{1}{2}\;\left| {\vec a \times \;\vec b} \right|\)= \(\frac{{\sqrt {389} }}{2}\)

Concept:

Area of a parallelogram with vectors \(\rm \vec {d_{1}}\) and \(\rm \vec {d_{1}}\) as its diagonals is given by: \(\rm Area=\dfrac{1}{2}\left|\vec{d_1}\times\vec{d_2}\right|\).

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:

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

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:

The given diagonals of the parallelogram are \(\rm \vec{a}= 3\hat{i} + \hat{j} - 2\hat{k}\) and \(\rm \vec{b}=\hat{i}-3\hat{j}+4\hat{k}\).

Using the formula for the area of a parallelogram whose diagonals \(\rm \vec {a}\) and \(\rm \vec {b}\) are given, we get:

\(\rm Area=\dfrac{1}{2}\left|\vec a\times\vec b\right|=\dfrac{1}{2}\left|(a_2b_3-a_3b_2)\hat i+(a_3b_1-a_1b_3)\hat j+(a_1b_2-a_2b_1)\hat k\right|\)

= \(\rm \dfrac{1}{2}\sqrt{(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2}\)

= \(\rm \dfrac{1}{2}\sqrt{[(1)(4)-(-2)(-3)]^2+[(-2)(1)-(3)(4)]^2+[(3)(-3)-(1)(1)]^2}\)

= \(\rm \dfrac{1}{2}\sqrt{(4-6)^2+(-2-12)^2+(-9-1)^2}\)

= \(\rm \dfrac{1}{2}\sqrt{4+196+100}\)

= \(\rm \dfrac{1}{2}\sqrt{300}\)

= \(5\sqrt{3}\).

Additional Information

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

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:

• Diagonals of a parallelogram bisect each other.

Calculation:

Since, the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both.

\(\Rightarrow \frac{{\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OC}}} }}{2} = \overrightarrow {{\rm{OP}}} \) 

\(\therefore \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OC}}} = 2 \times \overrightarrow {{\rm{OP}}} \)         …. (1)

Now,

\( \Rightarrow \frac{{\overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OD}}} }}{2} = \overrightarrow {{\rm{OP}}} \)  

\(\therefore \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OD}}} = 2 \times \overrightarrow {{\rm{OP}}} \)         …. (2)

Adding equation 1 and 2, we get

\(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} = 4{\rm{\;}}\overrightarrow {{\rm{OP}}} {\rm{\;}}\) 

Concept:

 A triangle ABC is said to be an isosceles triangle if triangle ABC must have two sides of equal length.

Calculations:

Given, the position vector of a point P with respect to origin O is î + 3ĵ - 2k̂ and that of a point Q is 3î + ĵ - 2k̂.

⇒\(\rm \bar {OP} \) = î + 3ĵ - 2k̂ and  \(\rm \bar {OQ}\) = 3î + ĵ - 2k̂.

⇒ |OP| = \(\rm \sqrt {1+9 + 4} = \sqrt {14}\)

⇒ |OQ| = \(\rm \sqrt {9 + 1+ 4} = \sqrt {14}\)

Here, |OP| = |OQ|

\(\rm \triangle POQ\) is isoscale.

The position vector of the bisector of the angle POQ = \(\rm \dfrac 1 2 (OP + OQ)\)

⇒The position vector of the bisector of the angle POQ = \(\rm \dfrac 1 2 [(î + 3ĵ - 2k̂) + (3î + ĵ - 2k̂)]\)

⇒The position vector of the bisector of the angle POQ = \(\rm \dfrac 1 2 (4î + 4ĵ - 4k̂) \)

⇒The position vector of the bisector of the angle POQ = \(\rm 2î + 2ĵ - 2k̂\)

Concept:

Considering a parallelogram ABCD, AC and BD are the diagonals that bisect each other at O.

We know that, the diagonal of the parallelogram bisects the parallelogram into two triangles of equal area.

Area of the parallelogram = 2 × area of ∆BCD.

In ∆BCD,

Base = BD and Height = CE = OC × sin θ = ½ × AC × sin θ

Area of triangle BCD = ½ × base × height = 1/2 × |\(\overrightarrow {BD} \)| × |\(\frac{{\overrightarrow {AC} }}{2}\) sin θ|

So, area of the parallelogram ABCD = |\(\overrightarrow {BD} \)| × |\(\frac{{\overrightarrow {AC} }}{2}\) sin θ | = 1/2 × |\(\overrightarrow {BD} \times \overrightarrow {AC} \)|

Calculation:

Given:

Let us assume the diagonals AC and BD as,

\(\overrightarrow {AC} \) = 3î + ĵ - 2k̂

\(\overrightarrow {BD}\) = î - 3ĵ + 4k̂

To find: Area of the parallelogram?

Area of the parallelogram = ½ × |\(\overrightarrow {BD} \times \overrightarrow {AC} \)|

= ½ × \(\left| {\begin{array}{*{20}{c}} i&j&k\\ 3&1&-2\\ 1&-3&4 \end{array}} \right|\)

= ½ × |î {4 – 6} ĵ – {12 – (-2)} + k̂ {-9 – 1}|

= ½ × |-2î - 14ĵ – 10 k̂|

= ½ × \(\sqrt {{2^2} + {{14}^2} + {{10}^2}} \)

= ½ × √(4 + 196 + 100)

= ½ × √(300)

= ½ × 10√3

= 5√3

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.

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

Calculation:

Given vectors are \(\rm 3\hat{i} - 8\hat{j} + 5\hat{k}, \hat{i} - λ\hat{j} + 2\hat{k} \;and\; 2\hat{i} + 3\hat{j} + 3\hat{k}\)

Using triangle law of vector addition,

\(\rm ⇒ 3\hat{i} - 8\hat{j} + 5\hat{k} = (\hat{i} - λ\hat{j} + 2\hat{k}) + (2\hat{i} + 3\hat{j} + 3\hat{k})\)

\(\rm ⇒ 3\hat{i} - 8\hat{j} + 5\hat{k} = 3\hat{i} + (- λ+3)\hat{j} + 5\hat{k} \)

Comparing the coefficient of \(\rm \vec {j}\), we get

⇒ -8 = -λ + 3

⇒ λ = 3 + 8

∴ λ = 11

Concept:

I. If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) then \(\vec a \times \;\vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right|\)

II. If \(\vec a\;and\;\vec b\) are the adjacent sides of a triangle, then area of triangle is given by: \(\frac{1}{2}\;\left| {\vec a \times \;\vec b} \right|\)

Calculation:

Given: In ΔOAB, where O is the origin, \(\overrightarrow {OA} = \;3\hat i - \hat j + \hat k\;and\;\overrightarrow {OB} = \;2\hat i + \hat j - 3\hat k\)

\(\overrightarrow {OA} \times \;\overrightarrow {OB} = \;\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 3&{ - 1}&1\\ 2&1&{ - 3} \end{array}} \right| = 2\;\hat i + 11\;\hat j + 5\;\hat k\)

\( \Rightarrow \left| {\overrightarrow {OA} \times \;\overrightarrow {OB} } \right| = \sqrt {{2^2} + {{11}^2} + {5^2}} = 5\sqrt 6 \)

As we know that, If \(\vec a\;and\;\vec b\) are the adjacent sides of a triangle, then area of triangle is given by: \(\frac{1}{2}\;\left| {\vec a \times \;\vec b} \right|\)

\( \Rightarrow \frac{1}{2}\left| {\overrightarrow {OA} \times \;\overrightarrow {OB} } \right| = \frac{1}{2} \times \;5\sqrt 6 = \frac{{5\sqrt 6 }}{2}\;sq\;units\)

Concept:

Area of triangle when two vectors are given:

\(\rm \frac12 \times |\vec {AB} \times \vec {AC}| \)

Cross product:

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

Calculation:

Here, Let A = (0,2,2), B = (2,0,-1) and C =(3,4,0) 

AB = (2-0, 0-2, -1-2) = (2, -2, -3) and

AC = (3-0, 4-2, 0-2) = (3, 2, -2)

Area of triangle = \(\rm \frac12 \times |\vec {AB} \times \vec {AC}| \)

 \(\begin{array}{l} =\rm \frac{1}{2}\left|\begin{array}{ccc} \rm \hat i &\rm \hat j & \rm \hat k \\ 2 & -2 & -3 \\ 3 & 2 & -2 \end{array}\right| \\ =\frac{1}{2}|[\rm \hat i(4+6)+ \rm \hat j(-4+9)+ \rm \hat k(4+6)]| \end{array}\)

=1/2(10 i + 5 j + 10 k)

= 1/2 √(100 + 25 + 100)

= 15/2 sq unit

Hence, option (1) is correct.

From question, the vectors \({\rm{\vec a\;and\;\vec b}}\) are non-collinear.

Then, we can write,

\(\Rightarrow {\rm{\vec a}} \neq \lambda {\rm{\vec b}}\)

for some non-zero scalar λ.

From question,

\(\vec \alpha = \left( {\lambda - 2} \right){\rm{\vec a}} + {\rm{\vec b}}\)

\(\vec \beta = \left( {4\lambda - 2} \right)\vec a + 3\vec b\)

So, we can write,

\(\vec \alpha = k\vec \beta \) for some k ∈ R -{0}

On substituting the values,

\(\Rightarrow \left( {\lambda - 2} \right){\rm{\vec a}} + {\rm{\vec b}} = k\left[ {\left( {4\lambda - 2} \right)\vec a + 3\vec b} \right]\)

\(\Rightarrow \left[ {\left( {\lambda - 2} \right) - k\left( {4\lambda - 2} \right)} \right]{\rm{a}} + \left( {1 - 3k} \right){\rm{b}} = 0\)

From question, as \({\rm{\vec a\;and\;\vec b}}\) are non-collinear, therefore they are linearly independent.

⇒ (λ - 2) - k(4λ - 2) = 0 and (1 - 3k) = 0

Now,

⇒ 1 = 3k

\(\therefore k = \frac{1}{3}\)

On substituting value of ‘k’ in another obtained equation,

\(\Rightarrow \left( {\lambda - 2} \right) - \frac{1}{3}\left( {4\lambda - 2} \right) = 0\)

⇒ 3λ - 6 = 4λ - 2

Concept:

The area of the quadrilateral ABCD = \(\rm \dfrac 12 |\vec {AC} \times \vec {BD}|\)  , where \(\rm \vec {AC} \;\; \text {and }\;\; \vec {BD}\) are diagonals.

Calculations:

Let \(\overrightarrow{AC}=2\hat{i}+\hat{j}+\hat{k}\)    and  \(\overrightarrow{BD}=-\hat{i}+3\hat{j}+2\hat{k}\) are the diagonal of the quadrilateral ABCD.

The area of the quadrilateral ABCD = \(\rm \dfrac 12 |\vec {AC} \times \vec {BD}|\) ....(1) 

⇒\(\rm (\vec {AC} \times \vec {BD}) =\)\(\begin{vmatrix} \vec i&\vec j & \vec k \\ 2&1 &1 \\ -1&3 &2 \end{vmatrix}\)

⇒\(\rm (\vec {AC} \times \vec {BD}) =\)\(\rm -\vec i - 5\vec j + 7 \vec k\)

⇒\(\rm |\vec {AC} \times \vec {BD}| = \sqrt {(-1)^2+(-5)^2+(7)^2}\)

⇒\(\rm |\vec {AC} \times \vec {BD}| = \sqrt {75} = 5 \sqrt 3\)

From equation (1), we have

The area of the quadrilateral ABCD = \(\rm \dfrac{5\sqrt3}{2}\)