CBSE 12th Standard Maths Subject Vector Algebra Case Study Questions 2021
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CBSE 12th Standard Maths Vector Algebra Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Maths
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Ginni purchased an air plant holder which is in the shape of a tetrahedron.
Let A, B, C and D are the coordinates of the air plant holder where A \(\equiv \) (1, 1, 1), B \(\equiv \) (2, 1, 3), C \(\equiv \) (3, 2, 2) and D \(\equiv \)(3, 3, 4).
Based on the above information, answer the following questions.
(i) Find the position vector of \(\overrightarrow{A B} \).(a) \(-\hat{i}-2 \hat{k}\) (b) \(2 \hat{i}+\hat{k}\) (c) \(\hat{i}+2 \hat{k}\) (d)\(-2 \hat{i}-\hat{k}\) (ii) Find the position vector of \(\overrightarrow{A C} \).
(a) \(2 \hat{i}-\hat{j}-\hat{k}\) (b) \(2 \hat{i}+\hat{j}+\hat{k}\) (c) \(-2 \hat{i}-\hat{j}+\hat{k}\) (d) \(\hat{i}+2 \hat{j}+\hat{k}\) (iii) Find the position vector of \(\overrightarrow{AD} .\).
(a) \( 2 \hat{i}-2 \hat{j}-3 \hat{k}\) (b) \( \hat{i}+\hat{j}-3 \hat{k}\) (c) \(3 \hat{i}+2 \hat{j}+2 \hat{k}\) (d) \(2\hat{i}+2 \hat{j}+3 \hat{k}\) (iv) Area of \(\Delta A B C\) =
(a) \(\frac{\sqrt{11}}{2} \mathrm{sq .units}\) (b) \(\frac{\sqrt{14}}{2} sq. units\) (c) \(\frac{\sqrt{13}}{2}\) (d)\(\frac{\sqrt{17}}{2} \mathrm{sq .units}\) (v) Find the unit vector along \(\overrightarrow{AD} .\)
(a) \(\frac{1}{\sqrt{17}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\) (b)\(\frac{1}{\sqrt{17}}(3 \hat{i}+3 \hat{j}+2 \hat{k})\) (c) \(\frac{1}{\sqrt{11}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\) (d) \((2 \hat{i}+2 \hat{j}+3 \hat{k})\) -
Teams A, B, C went for playing a tug of war game. Teams A, B, C have attached a rope to a metal ring and is trying to pull the ring into their own area (team areas shown below).
Team A pulls with force F1 = \(\hat{4}+\hat{0} \hat{j}\) KN
Team B ⟶ F2 = \(-2 \hat{i}+4 \hat{j}\) KN
Team C ⟶ F3 = \(-3 \hat{i}-3 \hat{j}\) KN
Based on the above information, answer the following questions.
(i) Which team will win the game ?(a) Team B (b) Team A (c) Team C (d) No one (ii) What is the magnitude of the teams combined force ?
(a) 7 KN (b) 1.4 KN (c) 1.5 KN (d) 2 KN (iii) In what direction is the ring getting pulled?
(a) 2.0 radian (b) 2.5 radian (c) 2.4 radian (d) 3 radian (iv) What is the magnitude of the force of Team B?
(a) 2\(\sqrt 5\) KN (b) 6 KN (c) 2 KN (d) \(\sqrt 6\) KN (v) How many KN force is applied by Team A?
(a) 5 KN (b) 4 KN (c) 2 KN (d) 16 KN -
Ishaan left from his village on weekend. First, he travelled up to temple. After this, he left for the zoo. After this he left for shopping in a mall. The positions of Ishaan at different places is given in the following graph.
Based on the above information, answer the following questions.
(i) Position vector of B is(a) \(3 \hat{i}+5 \hat{j}\) (b) \(5 \hat{i}+3 \hat{j}\) (c) \(-5 \hat{i}-3 \hat{j}\) (d) \(-5 \hat{i}+3 \hat{j}\) (ii) Position vector of D is
(a) \(5 \hat{i}+3 \hat{j}\) (b) \(3 \hat{i}+5 \hat{j}\) (c) \(8 \hat{i}+9 \hat{j}\) (d) \(9 \hat{i}+8 \hat{j}\) (iii) Find the vector \(\overrightarrow{B C}\) in terms of \(\hat{i}, \hat{j}\).
(a) \( \hat{i}-2 \hat{j}\) (b) \( \hat{i}+2 \hat{j}\) (c) \(2\hat{i}+ \hat{j}\) (d) \(2\hat{i}- \hat{j}\) (iv) Length of vector \(\overrightarrow{A D}\) is
(a) \(\sqrt 67\) units (b) \(\sqrt 85\) units (c) 90 units (d) 100 units (v) If \(\vec{M}=4 \hat{\jmath}+3 \hat{k}\) , then its unit vector
(a) \(\frac{4}{5} \hat{j}+\frac{3}{5} \hat{k}\) (b) \(\frac{4}{5} \hat{j}-\frac{3}{5} \hat{k}\) (c) \(-\frac{4}{5} \hat{j}+\frac{3}{5}\hat{k}\) (d) \(-\frac{4}{5} \hat{j}-\frac{3}{5} \hat{k}\) -
If two vectors are represented by the two sides of a triangle taken in order, then their sum is represented by the third side of the triangle taken in opposite order and this is known as triangle law of vector addition. lased on the above information, answer the following questions.
(i) If \(\vec{p}, \vec{q}, \vec{r}\)are the vectors represented by the sides of a triangle taken in order, then \( \vec{q} +\vec{r}\) =(a) \(\vec{p}\) (b) \(2 \vec{p}\) (c) \(-\vec{p}\) (d) None of these (ii) If ABCD is a parallelogram and AC and BD are its diagonals, then\( \vec{AC} +\vec{BD}\) =
(a) \(2 \vec{DA}\) (b) \(2 \vec{AB}\) (c) \(2\overrightarrow{BC}\) (d) \(2\vec{BD}\) (iii) If ABCD is a parallelogram, where \(\overrightarrow{A B}\)\(=2\overrightarrow{a}\) and \(\overrightarrow{BC}\) \(=2\overrightarrow{b}\), then \( \vec{AC} -\vec{BD}\) =
(a) \(3\vec{a}\) (b) \(4\vec{a}\) (c) \(2\vec{b}\) (d) \(4\vec{b}\) (iv) If ABCD is a quadrilateral whose diagonals are \( \vec{AC}\) and \(\vec{BD}\), then \( \vec{BA} +\vec{CD}\) =
(a) \(\overrightarrow{A C}+\overrightarrow{D B}\) (b) \(\overrightarrow{A C}+\overrightarrow{B D}\) (c) \(\overrightarrow{B C}+\overrightarrow{A D}\) (d) \(\overrightarrow{B D}+\overrightarrow{C A}\) (v) If T is the mid point of side YZ of \(\triangle\)XYZ, then\(\overrightarrow{XY}\) + \(\overrightarrow{XZ}\) =
(a) \(2\vec{YT}\) (b) \(2\vec{XT}\) (c) \(2\vec{TZ}\) (d) None of these -
A barge is pulled into harbour by two tug boats as shown in the figure.
Based on the above information, answer the following questions.
(i) Position vector of A is(a) \(4 \hat{i}+2 \hat{j}\) (b) \(4 \hat{i}+10 \hat{j}\) (c)\(4 \hat{i}-10 \hat{j}\) (d) \(4 \hat{i}-2 \hat{j}\) (ii) Position vector of B is
(a) \(4 \hat{i}+4 \hat{j}\) (b) \(6 \hat{i}+6 \hat{j}\) (c) \( 9 \hat{i}+7 \hat{j}\) (d) \(3 \hat{i}+3 \hat{j}\) (iii) Find the vector \(\vec{AC}\) in terms of \(\hat{i}, \hat{j}\)
(a) \(8 \hat{j}\) (b) \(-8 \hat{j}\) (c) \(8 \hat{i}\) (d) None of these (iv) If \(\vec{A}=\hat{i}+2 \hat{j}+3 \hat{k}\), then its unit vector is
(a)\(\frac{\hat{i}}{\sqrt{14}}+\frac{2 \hat{j}}{\sqrt{14}}+\frac{3 \hat{k}}{\sqrt{14}}\) (b) \(\frac{3 \hat{i}}{\sqrt{14}}+\frac{2 \hat{j}}{\sqrt{14}}+\frac{\hat{k}}{\sqrt{14}}\) (c) \(\frac{2 \hat{i}}{\sqrt{14}}+\frac{3 \hat{j}}{\sqrt{14}}+\frac{\hat{k}}{\sqrt{14}}\) (d) None of these (v) If \(\vec{A}=4 \hat{i}+3 \hat{j}\) and \(\vec{B}=3 \hat{i}+4 \hat{j}\), then IAI+ IBI = ___________.
(a) 12 (b) 13 (c) 14 (d) 10
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CBSE 12th Standard Maths Vector Algebra Case Study Questions 2021 Answer Keys
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(i) (c): Position vector of \(\overrightarrow{A B} \)
\(=(2-1) \hat{i} \dot{+}(1-1) \hat{j}+(3-1) \hat{k}=\hat{i}+2 \hat{k}\)
(ii) (b): Position vector of \(\overrightarrow{A C} \)
\(=(3-1) \hat{i}+(2-1) \hat{j}+(2-1) \hat{k}=2 \hat{i}+\hat{j}+\hat{k}\)
(iii) (d): Position vector of \(\overrightarrow{AD} \)
\(=(3-1) \hat{i}+(3-1) \hat{j}+(4-1) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
(iv) (b): Area of \(\Delta A B C\) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{array}\right|=\hat{i}(0-2)-\hat{j}(1-4)+\hat{k}(1-0)\)
\(=-2 \hat{i}+3 \hat{j}+\hat{k}\)
\( \Rightarrow |\overrightarrow{A B} \times \overrightarrow{A C}| =\sqrt{(-2)^{2}+3^{2}+1^{2}} \)
\(=\sqrt{4+9+1}=\sqrt{14}\)
Area of \(\Delta A B C\) \(=\frac{1}{2} \sqrt{14} \text { sq. units }\)
(v) (a): Unit vector along \(\overrightarrow{A D}=\frac{\overrightarrow{A D}}{|\overrightarrow{A D}|}\)
\(=\frac{2 \hat{i}+2 \hat{j}+3 k}{\sqrt{2^{2}+2^{2}+3^{2}}}=\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{4+4+9}}=\frac{1}{\sqrt{17}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\) -
Here, \(\left|\vec{F}_{1}\right|=\sqrt{(4)^{2}+0^{2}}=4 \mathrm{KN}\)
\( \left|\vec{F}_{2}\right|=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20} \mathrm{KN} \)
\(\left|\vec{F}_{3}\right|=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18} \mathrm{KN}\)
(i) (a): Since, \(\sqrt 20\) is larger. So, team B will win the game.
(ii) (b): Let F be the combined force
\(\therefore \vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=4 \hat{i}+0 \hat{j}-3 \hat{i}-3 \hat{j}-2 \hat{i}+4 \hat{j} \)
\(=-\hat{i}+\hat{j} \)
\(\therefore |\vec{F}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}=1.4 \mathrm{KN}\)
(iii) (c) : We have, \(\vec{F}=-\hat{i}+\hat{j}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=\tan ^{-1}\left(\frac{1}{-1}\right)=\frac{3 \pi}{4} \text { radian }\)
= 0.75 x 3.14 radian = 2.3555 radian ≈ 2.4 radian
(iv) (a): Magnitude of force of Team B = \(\sqrt 20\) KN
= 2\(\sqrt 5\) KN
(v) (b): 4 KN force is applied by team A. -
(i) (b): Here (5, 3) are the coordinates of B.
\(\therefore\) P.V. of B = \(5 \hat{i}+3 \hat{j}\)
(ii) (d) : Here (9, 8) are the coordinates of D.
\(\therefore\) P.V. of D = \({9} \hat{i}+8 \hat{j}\)
(iii) (b) : P.V. of B = \(5 \hat{i}+3 \hat{j}\) and ஃ P.V. of C = \(6 \hat{i}+5 \hat{j}\)
\(\therefore \ \overrightarrow{B C}=(6-5) \hat{i}+(5-3) \hat{j}=\hat{i}+2 \hat{j}\)
(iv) (b): Since P.V. of A = \(2 \hat{i}+2 \hat{j}\) , P.V. of D = \({9} \hat{i}+8 \hat{j}\)
\( \therefore \overrightarrow{A D}=(9-2) \hat{i}+(8-2) \hat{j}=7 \hat{i}+6 \hat{j} \)
\(|\overrightarrow{A D}|^{2}=7^{2}+6^{2}=49+36=85 \)
\(\Rightarrow |\overrightarrow{A D}|=\sqrt{85} \text { units }\)
(v) (a): We have, \(\vec{M}=4 \hat{j}+3 \hat{k}\) ,
\( \therefore |\vec{M}|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5 \)
\(\therefore \hat{M}=\frac{\vec{M}}{|\vec{M}|}=\frac{4 \hat{j}+3 \hat{k}}{5}=\frac{4}{5} \hat{j}+\frac{3}{5} \hat{k}\) -
(i) (c) : Let OAB be a triangle such that
\(\overrightarrow{A O}=-\vec{p}, \overrightarrow{A B}=\vec{q}, \)\(\overrightarrow{B O}=\vec{r}\)
Now, \(\vec{q}+\vec{r} =\overrightarrow{A B}+\overrightarrow{B O}\)
\(=\overrightarrow{A O} =-\vec{p}\)
(ii) (c) : From triangle law of vector addition,
\(\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{B C}+\overrightarrow{C D}\)
\( =\overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C D}\)
\(=\overrightarrow{A B}+2 \overrightarrow{B C}-\overrightarrow{A B}=2 \overrightarrow{B C} \)
(iii) (b) : \(\operatorname{In} \Delta A B C, \overrightarrow{A C}=2 \vec{a}+2 \vec{b}\)
and in \(\Delta A B D, 2 \vec{b}=2 \vec{a}+\overrightarrow{B D}\)..(ii)
[By triangle law of addition]
Adding (i) and (ii), we have
\( \overrightarrow{A C}+2 \vec{b}=4 \vec{a}+\overrightarrow{B D}+2 \vec{b} \)
⇒\( \overrightarrow{A C}-\overrightarrow{B D}=4 \vec{a} \)
(iv) (d) : \(\text { In } \Delta A B C, \overrightarrow{B A}+\overrightarrow{A C}=\overrightarrow{B C}\)... (i)
[By triangle law]
In \(\Delta B C D, \overrightarrow{B C}+\overrightarrow{C D}=\overrightarrow{B D}\) ..(i)
From (i) and (ii), \(\overrightarrow{B A}+\overrightarrow{A C}=\overrightarrow{B D}-\overrightarrow{C D}\)
\(\Rightarrow \overrightarrow{B A}+\overrightarrow{C D}=\overrightarrow{B D}-\overrightarrow{A C}=\overrightarrow{B D}+\overrightarrow{C A}\)
(v) (b): Since T is the mid point of YZ.
So,\(\overrightarrow{Y T}=\overrightarrow{T Z}\)
Now,\(\overrightarrow{X Y}+\overrightarrow{X Z}=(\overrightarrow{X T}+\overrightarrow{T Y})+(\overrightarrow{X T}+\overrightarrow{T Z})\)
[By triangle law]
\(=2 \overrightarrow{X T}+\overrightarrow{T Y}+\overrightarrow{T Z}=2 \overrightarrow{X T} \quad[\because \overrightarrow{T Y}=-\overrightarrow{Y T}]\) -
(i) (b): Here, (4, 10) are the coordinates of A.
\(\therefore\) P.V. of A = \(4 \hat{i}+10 \hat{j}\)
(ii) (c): Here, (9, 7) are the coordinates of B.
\(\therefore\) P.V. of B \(=9 \hat{i}+7 \hat{j}\)
(iii) (b): Here, P.V. of A = \(4 \hat{i}+10 \hat{j}\) and P.V. of \(C=4 \hat{i}+2 \hat{j}\)
\(\therefore \ \overrightarrow{A C}=(4-4) \hat{i}+(2-10) \hat{j}=-8 \hat{j}\)
(iv) (a): Here \(\vec{A}=\hat{i}+2 \hat{j}+3 \hat{k}\)
\( \therefore \ |\vec{A}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \)
\(\therefore \ \hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)
(v) (d): We have, \(\vec{A}=4 \hat{i}+3 \hat{j}\) and \( \vec{B}=3 \hat{i}+4 \hat{j}\)
\(\therefore|\vec{A}|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5\)
and \(|\vec{B}|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5\)
Thus, \(|\vec{A}|+|\vec{B}|=5+5=10\)
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