CBSE Class 12 Mathematics Vector Algebra Case Studies

Vector Algebra

Read the following and answer any four questions. Solar Panels have to be installed carefully so that the tilt of the roof, and the direction to the sun, produce the largest possible electrical power in the solar panels. A surveyor uses his instrument to determine the coordinates of the four corners of a roof where solar panels are to be mounted. In the picture , suppose the points are labelled counter clockwise from the roof corner nearest to the camera in units of meters \( P_1 (6,8,4) \), \( P_2 (21,8,4) \), \( P_3 (21,16,10) \) and \( P_4 (6,16,10) \).

Question. What are the components to the two edge vectors defined by \( \vec{A} = PV \text{ of } P_2 - PV \text{ of } P_1 \) and \( \vec{B} = PV \text{ of } P_4 - PV \text{ of } P_1 \)? (where \( PV \) stands for position vector)
(a) (15, 8, 4), (0, 8, 6)
(b) (15, 0, 0), (6, 8, 0)
(c) (15, 0, 0), (0, 8, 6)
(d) (15, 8, 4), (6, 8, 0)
Answer: (c)

Question. The vector in standard notation with \( \hat{i}, \hat{j} \) and \( \hat{k} \), (where \( \hat{i}, \hat{j} \) and \( \hat{k} \) are the unit vectors along the three axes) are
(a) \( (15\hat{i} + 8\hat{j} + 4\hat{k}), (0\hat{i} + 8\hat{j} + 6\hat{k}) \)
(b) \( (15\hat{i} + 0\hat{j} + 0\hat{k}), (6\hat{i} + 8\hat{j} + 6\hat{k}) \)
(c) \( (15\hat{i} + 0\hat{j} + 0\hat{k}), (0\hat{i} + 8\hat{j} + 6\hat{k}) \)
(d) \( (15\hat{i} + 8\hat{j} + 4\hat{k}), (6\hat{i} + 8\hat{j} + 6\hat{k}) \)
Answer: (c)

Question. What are the magnitudes of the vectors \( \vec{A} \) and \( \vec{B} \)?
(a) \( \sqrt{325} \) units, 10 units
(b) 15 units, \( \sqrt{136} \) units
(c) 15 units, 10 units
(d) \( \sqrt{325} \) units, \( \sqrt{136} \) units
Answer: (c)

Question. What are the components to the vector \( \vec{N} \), perpendicular to \( \vec{A} \) and \( \vec{B} \) and the surface of the roof?
(a) 0, –90, 120
(b) 0, –90, –120
(c) 0, 90, 120
(d) 0, 90, –120
Answer: (a)

Question. The sun is located along the unit vector \( \vec{S} = \frac{1}{2}\hat{i} - \frac{6}{7}\hat{j} + \frac{1}{7}\hat{k} \). If the flow of solar energy is given by the vector \( \vec{F} = 910 \vec{S} \) units of watts/meter\(^2\), what is the dot product of vectors \( \vec{F} \) with \( \vec{N} \), and the units for this quantity?
(a) 84,800 watts
(b) 85,800 watts
(c) 54600 watts
(d) 86255 watts
Answer: (b)

Read the following and answer any four questions. A class XII student appearing for a competitive examination was asked to attempt the following questions. Let \( \vec{a}, \vec{b} \) and \( \vec{c} \) be three non zero vectors.

Question. If \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \) then
(a) \( \vec{a} \perp \vec{b} \)
(b) \( \vec{a} \parallel \vec{b} \)
(c) \( \vec{a} = \vec{b} \)
(d) None of these
Answer: (a)

Question. If \( \vec{a} = \hat{i} - 2\hat{j}, \vec{b} = 2\hat{i} + \hat{j} + 3\hat{k} \) then evaluate \( (2\vec{a} + \vec{b}) \cdot [(\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b})] \)
(a) 0
(b) 4
(c) 3
(d) 2
Answer: (a)

Question. If \( \vec{a} \) and \( \vec{b} \) are unit vectors and \( \theta \) be the angle between them then \( |\vec{a} - \vec{b}| \) is
(a) \( \sin \frac{\theta}{2} \)
(b) \( 2 \sin \frac{\theta}{2} \)
(c) \( 2 \cos \frac{\theta}{2} \)
(d) \( \cos \frac{\theta}{2} \)
Answer: (b)

Question. Let \( \vec{a}, \vec{b} \) and \( \vec{c} \) be unit vectors such that \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0 \) and angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \) then \( \vec{a} = \)
(a) \( 2(\vec{b} \times \vec{c}) \)
(b) \( -2(\vec{b} \times \vec{c}) \)
(c) \( \pm 2(\vec{b} \times \vec{c}) \)
(d) \( \oplus (\vec{b} \pm \vec{c}) \)
Answer: (c)

Question. The area of the parallelogram formed by \( \vec{a} \) and \( \vec{b} \) as diagonals is
(a) 70
(b) 35
(c) \( \frac{\sqrt{70}}{2} \)
(d) \( \sqrt{70} \)
Answer: (c)

Read the following and answer any four questions. A cricket match is organized between two Clubs A and B for which a team from each club is chosen. Remaining players of Club A and Club B are respectively sitting on the plane represented by the equation \( \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3 \) and \( \vec{r} \cdot (\hat{i} + 3\hat{j} + 2\hat{k}) = 8 \) to cheer the team of their own clubs.

Question. The Cartesian equation of the plane on which players of Club A are seated is
(a) \( 2x - y + z = 3 \)
(b) \( 2x - y + 2z = 3 \)
(c) \( 2x - y + z = -3 \)
(d) \( x - y + z = 3 \)
Answer: (a)

Question. The magnitude of the normal to the plane on which players of club B are seated, is
(a) \( \sqrt{15} \)
(b) \( \sqrt{14} \)
(c) \( \sqrt{17} \)
(d) \( \sqrt{20} \)
Answer: (b)

Question. The intercept form of the equation of the plane on which players of Club B are seated is
(a) \( \frac{x}{8} + \frac{y}{8/3} + \frac{z}{4} = 1 \)
(b) \( \frac{x}{8/5} + \frac{y}{8/3} + \frac{z}{3} = 1 \)
(c) \( \frac{x}{8} + \frac{y}{8/3} + \frac{z}{4} = 1 \)
(d) \( \frac{x}{8} + \frac{y}{7} + \frac{z}{2} = 1 \)
Answer: (a)

Question. Which of the following is a player of Club B?
(a) Player sitting at (1, 2, 1)
(b) Player sitting at (0, 1, 2)
(c) Player sitting at (1, 4, 1)
(d) Player sitting at (1, 1, 2)
Answer: (d)

Question. The distance of the plane, on which players of Club B are seated, from the origin is
(a) \( \frac{8}{\sqrt{14}} \) units
(b) \( \frac{6}{\sqrt{14}} \) units
(c) \( \frac{7}{\sqrt{14}} \) units
(d) \( \frac{9}{\sqrt{14}} \) units
Answer: (a)

Read the following and answer any four questions. The Indian coast guard, while patrolling, saw a suspicious boat with people. They were nowhere looking like fishermen. The coast guard were closely observing the movement of the boat for an opportunity to seize the boat. They observed that the boat is moving along a planar surface. At an instant of time, the coordinates of the position of the coast guard helicopter and the boat is (1, 3, 5) and (2, 5, 3) respectively.

Question. If the line joining the positions of the helicopter and the boat is perpendicular to the plane in which the boat moves, then the equation of the plane is
(a) \( -x + 2y - 2z = 6 \)
(b) \( x + 2y + 2z = 6 \)
(c) \( x + 2y - 2z = 6 \)
(d) \( x - 2y - 2z = 6 \)
Answer: (c)

Question. If the coast guard decide to shoot the boat at that given instant of time, then what is the distance (in meters) that the bullet has to travel?
(a) 5m
(b) 3m
(c) 6m
(d) 4m
Answer: (b)

Question. If the coast guard decides to shoot the boat at that given instant of time, when the speed of bullet is 36m/sec, then what is the time taken for the bullet to travel and hit the boat?
(a) \( \frac{1}{8} \) seconds
(b) \( \frac{1}{14} \) seconds
(c) \( \frac{1}{10} \) seconds
(d) \( \frac{1}{12} \) seconds
Answer: (d)

Question. At that given instant of time, the equation of line passing through the positions of the helicopter and boat is
(a) \( \frac{x-1}{1} = \frac{y-3}{2} = \frac{z-5}{-2} \)
(b) \( \frac{x-1}{2} = \frac{y+3}{1} = \frac{z-5}{-2} \)
(c) \( \frac{x+1}{-2} = \frac{y-3}{-1} = \frac{z-5}{-2} \)
(d) \( \frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+5}{2} \)
Answer: (a)

Question. At a different instant of time, the boat moves to a different position along the planar surface. What should be the coordinates of the location of the boat if the coast guard shoots the bullet along the line whose equation is \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{1} \) for the bullet to hit the boat?
(a) \( (\frac{8}{3}, \frac{19}{3}, \frac{-14}{3}) \)
(b) \( (\frac{8}{3}, \frac{-19}{3}, \frac{-14}{3}) \)
(c) \( (\frac{8}{3}, \frac{-19}{3}, \frac{14}{3}) \)
(d) \( (\frac{8}{3}, \frac{19}{3}, \frac{14}{3}) \)
Answer: (d)

Read the following and answer any four questions. The equation of motion of a missile are \( x = 3t, y = - 4t, z = t \), where the time ‘\( t \)’ is given in seconds, and the distance is measured in kilometres.

Question. What is the path of the missile?
(a) Straight line
(b) Parabola
(c) Circle
(d) Ellipse
Answer: (a)

Question. Which of the following points lie on the path of the missile?
(a) (6, 8, 2)
(b) (6, –8, –2)
(c) (6, –8, 2)
(d) (-6, –8, 2)
Answer: (c)

Question. At what distance will the rocket be from the starting point (0, 0, 0) in 5 seconds?
(a) \( \sqrt{550} \) kms
(b) \( \sqrt{650} \) kms
(c) \( \sqrt{450} \) kms
(d) \( \sqrt{750} \) kms
Answer: (b)

Question. If the position of rocket at a certain instant of time is (5, –8, 10), then what will be the height of the rocket from the ground? (The ground is considered as the \( xy \)–plane).
(a) 12 km
(b) 11 km
(c) 20 km
(d) 10 km
Answer: (d)

Question. At a certain instant of time, if the missile is above the sea level, where the equation of the surface of sea is given by \( 2x + y + 3z = 1 \) and the position of the missile at that instant of time is (1, 1, 2), then the image of the position of the rocket in the sea is
(a) \( (\frac{-9}{7}, \frac{-1}{7}, \frac{-10}{7}) \)
(b) \( (\frac{9}{7}, \frac{-1}{7}, \frac{-10}{7}) \)
(c) \( (\frac{-9}{7}, \frac{1}{7}, \frac{-10}{7}) \)
(d) \( (\frac{-9}{7}, \frac{-1}{7}, \frac{10}{7}) \)
Answer: (a)

Read the following and answer any four questions. Suppose the floor of a hotel is made up of mirror polished Salvatore stone. There is a large crystal chandelier attached to the ceiling of the hotel room. Consider the floor of the hotel room as a plane having the equation \( x - y + z = 4 \) and the crystal chandelier is suspended at the point (1, 0, 1).

Question. The direction ratios of the perpendicular from the point (1, 0, 1) to the plane \( x - y + z = 4 \) is
(a) (–1, –1, 1)
(b) (1, –1, –1)
(c) (–1, –1, –1)
(d) (1, –1, 1)
Answer: (d)

Question. The length of the perpendicular from the point (1, 0, 1) to the plane \( x - y + z = 4 \) is
(a) \( \frac{2}{\sqrt{3}} \) units
(b) \( \frac{4}{\sqrt{3}} \) units
(c) \( \frac{6}{\sqrt{3}} \) units
(d) \( \frac{8}{\sqrt{3}} \) units
Answer: (a)

Question. The equation of the perpendicular from the point (1, 0, 1) to the plane \( x - y + z = 4 \) is
(a) \( \frac{x-1}{2} = \frac{y}{1} = \frac{z-1}{-5} \)
(b) \( \frac{x-1}{-2} = \frac{y}{-1} = \frac{z-1}{2} \)
(c) \( \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{1} \)
(d) \( \frac{x-1}{2} = \frac{y}{-2} = \frac{z-1}{1} \)
Answer: (c)

Question. The equation of the plane parallel to the plane \( x - y + z = 4 \), which is at a unit distance from the point (1, 0, 1) is
(a) \( x - y + z + (2 - \sqrt{3}) = 0 \)
(b) \( x - y + z - (2 + \sqrt{3}) = 0 \)
(c) \( x - y + z + (2 + \sqrt{3}) = 0 \)
(d) Both (a) and (c)
Answer: (d)

Question. The direction cosine of the normal to the plane \( x - y + z = 4 \) is
(a) \( (\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}) \)
(b) \( (\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \)
(c) \( (\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \)
(d) \( (\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}) \)
Answer: (b)