Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.10

Choose the Most Suitable Answer From the Given Four Alternatives

Question 1. If \( \overline { a } \) and \( \overline {b} \) are parallel vectors, then \( [\overline { a }, \overline { c }, \overline { b }] \) is equal to
(a) 2
(b) -1
(c) 1
(d) 0
Answer: (d) 0
In simple words: When two vectors are parallel, their scalar triple product with any other vector becomes zero. This is because a parallel combination cannot form a volume with another vector.

🎯 Exam Tip: Remember that if two vectors in a scalar triple product are parallel or identical, the value of the product is always zero.

Question 2. If a vector \( \overline { a } \) lies in the plane of \( \overline { \beta } \) and \( \overline { y } \), then \( [\overline { a }, \overline { \beta }, \overline { y }] \) is equal to
(a) \( [\overline { a }, \overline { \beta }, \overline { y }]= 1 \)
(b) \( [\overline { a }, \overline { \beta }, \overline { y }]= -1 \)
(c) \( [\overline { a }, \overline { \beta }, \overline { y }]= 0 \)
(d) \( [\overline { a }, \overline { \beta }, \overline { y }]= 2 \)
Answer: (c) \( [\overline { a }, \overline { \beta }, \overline { y }]= 0 \)
In simple words: If one vector lies in the plane formed by two other vectors, then these three vectors are coplanar. When vectors are coplanar, the volume of the parallelepiped they form is zero, meaning their scalar triple product is zero.

🎯 Exam Tip: Coplanar vectors mean their scalar triple product is zero, as they don't enclose any volume.

Question 3. If \( \overline { a } \cdot \overline { b } = \overline { b } \cdot \overline { c } = \overline { c } \cdot \overline { a } = 0 \), then the value of \( [\overline { a }, \overline { b }, \overline { c }] \) is
(a) \( |\overline { a }| |\overline { b }| |\overline { c }| \)
(b) \( |\overline { a }| |\overline { b }| |\overline { c }| \)
(c) 1
(d) -1
Answer: (a) \( |\overline { a }| |\overline { b }| |\overline { c }| \)
In simple words: When the dot product of each pair of three vectors is zero, it means they are all perpendicular to each other. In this special case, the scalar triple product is simply the product of their magnitudes.

🎯 Exam Tip: If three vectors are mutually perpendicular, their scalar triple product is the product of their individual magnitudes. This is a common property for orthogonal vectors.

Question 4. If \( \overline { a }, \overline { b }, \overline { c } \) are three unit vectors such that \( \overline { a } \) is perpendicular to \( \overline {b} \) and is parallel to \( \overline { c } \) then \( \overline { a } \times (\overline { b } \times \overline { c }) \) is equal to
(a) \( \overline { a } \)
(b) \( \overline {b} \)
(c) \( \overline {c} \)
(d) \( \overline { 0 } \)
Answer: (b) \( \overline { b } \)
In simple words: If vector \( \overline { a } \) is perpendicular to \( \overline { b } \) and parallel to \( \overline { c } \), then using the vector triple product formula, the result simplifies to \( \overline { b } \). This happens because \( \overline {a} \cdot \overline{c} \) is 1 and \( \overline{a} \cdot \overline{b} \) is 0.

🎯 Exam Tip: Utilize the vector triple product formula \( \overline{A} \times (\overline{B} \times \overline{C}) = (\overline{A} \cdot \overline{C})\overline{B} - (\overline{A} \cdot \overline{B})\overline{C} \) and the properties of unit vectors for quick solutions.

Question 5. If \( [\overline { a }, \overline { b }, \overline {c} ] = 1 \) then the value of \( \frac{\overline{a} \cdot (\overline{b} \times \overline{c})}{(\overline{c} \times \overline{a}) \cdot \overline{b}} + \frac{\overline{b} \cdot (\overline{c} \times \overline{a})}{(\overline{a} \times \overline{b}) \cdot \overline{c}} + \frac{\overline{c} \cdot (\overline{a} \times \overline{b})}{(\overline{c} \times \overline{b}) \cdot \overline{a}} \) is
(a) 1
(b) -1
(c) 2
(d) 3
Answer: (a) 1
In simple words: The scalar triple product \( [\overline{a}, \overline{b}, \overline{c}] \) can be written in many equivalent ways by cyclic permutation. When \( [\overline{a}, \overline{b}, \overline{c}] = 1 \), each fraction in the expression simplifies due to these cyclic properties.

🎯 Exam Tip: Remember that the scalar triple product \( [\overline{a}, \overline{b}, \overline{c}] \) is invariant under cyclic permutation of vectors (e.g., \( [\overline{a}, \overline{b}, \overline{c}] = [\overline{b}, \overline{c}, \overline{a}] \)), which simplifies many expressions.

Question 6. The volume of the parallelepiped with its edges represented by the vectors \( \hat { i } + \hat { j } \), \( \hat { i } + 2\hat { j } \), \( \hat { i } + \hat {j} + \pi\hat { k } \) is
(a) \( \frac { \pi }{ 2 } \)
(b) \( \frac { \pi }{ 3 } \)
(c) \( \pi \)
(d) \( \frac { \pi }{ 4 } \)
Answer: (c) \( \pi \)
In simple words: To find the volume of a parallelepiped with given edge vectors, you calculate the absolute value of their scalar triple product. This involves finding the determinant of the matrix formed by their components.

🎯 Exam Tip: The volume of a parallelepiped formed by vectors \( \overline{u}, \overline{v}, \overline{w} \) is given by \( | \overline{u} \cdot (\overline{v} \times \overline{w}) | \), which is the absolute value of the determinant of the matrix formed by their components.

Question 7. If \( \overline { a } \) and \( \overline {b} \) are unit vectors such that \( [\overline { a }, \overline { b }, \overline { a } \times \overline { b }] = \frac { 1 }{ 4 } \), then the angle between \( \overline { a } \) and \( \overline {b} \) is
(a) \( \frac { \pi }{ 6 } \)
(b) \( \frac { \pi }{ 4 } \)
(c) \( \frac { \pi }{ 3 } \)
(d) \( \frac { \pi }{ 2 } \)
Answer: (a) \( \frac { \pi }{ 6 } \)
In simple words: The scalar triple product of \( \overline{a}, \overline{b}, \overline{a} \times \overline{b} \) can be rewritten using properties of the cross product. By equating this to the given value, we can find the sine of the angle between \( \overline{a} \) and \( \overline{b} \), which helps us find the angle itself.

🎯 Exam Tip: Use the identity \( [\overline{a}, \overline{b}, \overline{c}] = (\overline{a} \times \overline{b}) \cdot \overline{c} \). For \( \overline{c} = \overline{a} \times \overline{b} \), the expression becomes \( |\overline{a} \times \overline{b}|^2 \), simplifying calculations involving unit vectors.

Question 8. If \( \overline { a } = \hat { i } + \hat { j } + \hat { k }, \overline { b } = \hat { i } + \hat { j }, \overline { c } = \hat { i } \) and \( (\overline { a } \times \overline { b }) \times \overline { c } = \lambda\overline { a } + \mu\overline {b} \) then the value of \( \lambda + \mu \) is
(a) 0
(b) 1
(c) 6
(d) 3
Answer: (a) 0
In simple words: We use the vector triple product formula to simplify \( (\overline { a } \times \overline { b }) \times \overline { c } \). By comparing the result with \( \lambda\overline { a } + \mu\overline {b} \) and calculating the dot products of the given vectors, we find the values of \( \lambda \) and \( \mu \), and then their sum.

🎯 Exam Tip: For vector triple products like \( (\overline { A } \times \overline { B }) \times \overline { C } \), apply the formula \( (\overline { A } \cdot \overline { C })\overline { B } - (\overline { B } \cdot \overline { C })\overline { A } \). This directly gives the coefficients for \( \overline { A } \) and \( \overline { B } \).

Question 9. If \( \overline { a }, \overline { b }, \overline { c } \) are non-coplanar, non-zero vectors such that \( [\overline { a }, \overline { b }, \overline { c }] = 3 \), then \( \{[\overline { a } \times \overline { b }, \overline { b } \times \overline { c }, \overline { c } \times \overline { a }]^2\} \) is equal to
(a) 81
(b) 9
(c) 27
(d) 18
Answer: (a) 81
In simple words: There is a special identity that relates the scalar triple product of the vector products to the cube of the original scalar triple product. Since the original value is 3, its cube is 27, and then squaring that result gives 81.

🎯 Exam Tip: Remember the identity \( [\overline{a} \times \overline{b}, \overline{b} \times \overline{c}, \overline{c} \times \overline{a}] = [\overline{a}, \overline{b}, \overline{c}]^2 \). This simplifies calculations significantly.

Question 10. If \( \overline { a }, \overline { b }, \overline { c } \) are three non-coplanar vectors such that \( \overline { a } \times (\overline { b } \times \overline { c }) = \frac { \overline{b}+\overline{c} }{ \sqrt{2} } \) then the angle between \( \overline { a } \) and \( \overline {b} \) is
(a) \( \frac { \pi }{ 2 } \)
(b) \( \frac { 3\pi }{ 4 } \)
(c) \( \frac { \pi }{ 4 } \)
(d) \( \pi \)
Answer: (b) \( \frac { 3\pi }{ 4 } \)
In simple words: We use the vector triple product formula to expand the left side of the equation. By comparing the coefficients of \( \overline{b} \) and \( \overline{c} \) on both sides, we can find the dot product \( \overline{a} \cdot \overline{b} \), which then helps us determine the angle between \( \overline{a} \) and \( \overline{b} \).

🎯 Exam Tip: When dealing with vector equations, expand vector products using identities. Comparing coefficients of linearly independent vectors is a powerful technique to solve for unknowns or relationships.

Question 11. If the volume of the parallelepiped with \( \overline { a } \times \overline { b }, \overline { b } \times \overline { c }, \overline { c } \times \overline { a } \) as coterminous edges is 8 cubic units, then the volume of the parallelepiped with \( (\overline { a } \times \overline { b }) \times (\overline { b } \times \overline { c }) \), \( (\overline { b } \times \overline { c }) \times (\overline { c } \times \overline { a }) \) and \( (\overline { c } \times \overline { a }) \times (\overline { a } \times \overline { b }) \) as coterminous edges is
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Answer: (c) 64 cubic units
In simple words: We use the property that if the volume of a parallelepiped with edges \( \overline{u}, \overline{v}, \overline{w} \) is V, then the volume of the parallelepiped with edges \( \overline{u} \times \overline{v}, \overline{v} \times \overline{w}, \overline{w} \times \overline{u} \) is \( V^2 \). Applying this rule twice gives \( (V^2)^2 = V^4 \).

🎯 Exam Tip: Remember the identity: if \( V = [\overline{a}, \overline{b}, \overline{c}] \), then \( [\overline{a} \times \overline{b}, \overline{b} \times \overline{c}, \overline{c} \times \overline{a}] = V^2 \). Using this property saves significant computation time.

Question 12. Consider the vectors \( \overline { a }, \overline { b }, \overline { c }, \overline { d } \) such that \( (\overline { a } \times \overline { b }) \times (\overline { c } \times \overline { d }) = \overline { 0 } \). Let P\(_{1}\) and P\(_{2}\) be the planes determined by the pairs of vectors \( \overline { a }, \overline { b } \) and \( \overline { c }, \overline { d } \) respectively. Then the angle between P\(_{1}\) and P\(_{2}\) is
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Answer: (a) 0°
In simple words: The cross product \( \overline{a} \times \overline{b} \) gives a vector perpendicular to plane P\(_{1}\). Similarly, \( \overline{c} \times \overline{d} \) gives a vector perpendicular to plane P\(_{2}\). If their cross product is zero, it means these normal vectors are parallel, which implies the planes P\(_{1}\) and P\(_{2}\) are parallel. Parallel planes have an angle of 0° between them.

🎯 Exam Tip: The cross product of two vectors gives a normal vector to the plane containing them. If the cross product of two normal vectors is zero, the normal vectors are parallel, and thus the planes are parallel.

Question 13. If \( \overline { a } \times (\overline { b } \times \overline { c }) = (\overline { a } \times \overline { b }) \times \overline { c } \) where \( \overline { a }, \overline { b }, \overline { c } \) are any three vectors such that \( \overline { b } \cdot \overline { c } \neq 0 \) and \( \overline { a } \cdot \overline { b } \neq 0 \), then \( \overline { a } \) and \( \overline { c } \) are
(a) perpendicular
(b) parallel
(c) inclined at angle \( \frac { \pi }{ 3 } \)
(d) inclined at an angle \( \frac { \pi }{ 6 } \)
Answer: (b) parallel
In simple words: We expand both sides of the equation using the vector triple product formula. By simplifying the resulting equation, we find a relationship between the dot products \( \overline{a} \cdot \overline{c} \) and \( \overline{b} \cdot \overline{c} \). Since \( \overline{b} \cdot \overline{c} \neq 0 \) and \( \overline{a} \cdot \overline{b} \neq 0 \), this relationship implies that \( \overline{a} \) and \( \overline{c} \) must be parallel.

🎯 Exam Tip: Carefully apply the vector triple product identity: \( \overline{X} \times (\overline{Y} \times \overline{Z}) = (\overline{X} \cdot \overline{Z})\overline{Y} - (\overline{X} \cdot \overline{Y})\overline{Z} \). Comparing resulting vector expressions is key to finding relationships between vectors.

Question 14. If \( \overline { a } = 2\hat { i } + 3\hat { j } – \hat { k }, \overline { b } = \hat { i } + \hat { j } + \hat { k }, \overline { c } = 3\hat { i } + 5\hat { j } – \hat { k } \) then a vector perpendicular to \( \overline { a } \) and lies in the plane containing \( \overline {b} \) and \( \overline { c } \) is
(a) \( -17\hat {i } + 21\hat { j } – 97\hat { k } \)
(b) \( 17\hat {i } + 21\hat { j } – 123\hat { k } \)
(c) \( -17\hat { i } – 21\hat { j } + 97\hat { k } \)
(d) \( -17\hat {i } – 21\hat { j } – 97\hat { k } \)
Answer: (d) \( -17\hat {i } – 21\hat { j } – 97\hat { k } \)
In simple words: A vector that is perpendicular to \( \overline{a} \) and lies in the plane of \( \overline{b} \) and \( \overline{c} \) can be found by calculating the vector triple product \( \overline{a} \times (\overline{b} \times \overline{c}) \). First, find the cross product of \( \overline{b} \) and \( \overline{c} \), and then cross that result with \( \overline{a} \).

🎯 Exam Tip: To find a vector perpendicular to \( \overline{X} \) and coplanar with \( \overline{Y} \) and \( \overline{Z} \), calculate the vector triple product \( \overline{X} \times (\overline{Y} \times \overline{Z}) \). This operation produces a vector in the plane of \( \overline{Y} \) and \( \overline{Z} \) that is orthogonal to \( \overline{X} \).

Question 15. The angle between the lines \( \frac { x-2 }{ 3 } = \frac { y+1 }{ -2 }, z = 2 \) and \( \frac { x-1 }{ 1 } = \frac { 2y+3 }{ 3 } = \frac { z+5 }{ 2 } \) is
(a) \( \frac { \pi }{ 6 } \)
(b) \( \frac { \pi }{ 4 } \)
(c) \( \frac { \pi }{ 3 } \)
(d) \( \frac { \pi }{ 2 } \)
Answer: (d) \( \frac { \pi }{ 2 } \)
In simple words: First, find the direction vectors for both lines. The direction vector for the first line is \( \overline{n_1} = 3\hat{i} - 2\hat{j} + 0\hat{k} \). For the second line, rewrite its symmetric form to find the direction vector, which is \( \overline{n_2} = \hat{i} + \frac{3}{2}\hat{j} + 2\hat{k} \). Then, calculate their dot product. If the dot product is zero, the lines are perpendicular, meaning the angle between them is \( \frac{\pi}{2} \).

🎯 Exam Tip: To find the angle between two lines, use their direction vectors. The cosine of the angle \( \theta \) is given by \( \frac{|\overline{d_1} \cdot \overline{d_2}|}{|\overline{d_1}| |\overline{d_2}|} \). If the dot product is zero, the lines are perpendicular.

Question 16. If the line \( \frac { x-2 }{ 3 } = \frac { y-1 }{ -5 } = \frac { z+2 }{ 2 } \) lies in the plane \( x + 3y – \alpha z + \beta = 0 \) then \( (\alpha + \beta) \) is
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Answer: (b) (-6, 7)
In simple words: If a line lies in a plane, two conditions must be met: a point on the line must lie in the plane, and the direction vector of the line must be perpendicular to the normal vector of the plane. By using these two conditions, we can form equations to solve for \( \alpha \) and \( \beta \).

🎯 Exam Tip: For a line to lie in a plane, the dot product of the line's direction vector and the plane's normal vector must be zero, AND any point on the line must satisfy the plane's equation. These two conditions are necessary and sufficient.

Question 17. The angle between the line \( \overline { r } = (\hat { i } + 2\hat { j } – 3\hat { k }) + t(2\hat { i } + \hat { j } – 2\hat { k }) \) and the plane \( \overline { r } \cdot (\hat { i } + \hat {j}) + 4 = 0 \) is
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Answer: (c) 45°
In simple words: To find the angle between a line and a plane, we use the formula for the sine of the angle, \( \sin \theta = \frac{|\overline{b} \cdot \overline{n}|}{|\overline{b}| |\overline{n}|} \), where \( \overline{b} \) is the direction vector of the line and \( \overline{n} \) is the normal vector of the plane. After calculation, \( \sin \theta \) is found to be \( \frac{1}{\sqrt{2}} \), which corresponds to an angle of 45°.

🎯 Exam Tip: Be careful with the formula for the angle between a line and a plane; it uses \( \sin \theta \), unlike the angle between two lines or two planes which uses \( \cos \theta \).

Question 18. The co-ordinates of the point where the line \( \overline { r } = (6\hat { i } – \hat { j } – 3\hat { k }) + t(-\hat { i } + \hat { k }) \) meets the plane \( \overline { r } \cdot (\hat { i } + \hat { j } – \hat { k }) = 3 \) are
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Answer: (d) (5, -1, 1)
In simple words: Represent any point on the line using the parameter 't'. Substitute these coordinates into the plane's equation to find the value of 't'. Once 't' is known, plug it back into the line's equation to get the exact coordinates of the intersection point.

🎯 Exam Tip: To find the intersection point of a line and a plane, express a general point on the line in terms of a parameter (e.g., \( x = x_0 + at \)), substitute these expressions into the plane's equation, solve for the parameter, and then substitute the parameter value back into the line's equations to get the coordinates.

Question 19. Distance from the origin to the plane \( 3x – 6y + 2z + 7 = 0 \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
In simple words: The distance from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is given by the formula \( \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \). For the origin (0, 0, 0), this simplifies to \( \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \).

🎯 Exam Tip: Remember the formula for the perpendicular distance from a point to a plane. For the origin, it's simply the absolute value of the constant term divided by the magnitude of the normal vector.

Question 21. If the direction cosines of a line are \( \frac { 1 }{ c }, \frac { 1 }{ c }, \frac { 1 }{ c } \) then
(a) \( c = \pm 3 \)
(b) \( c = \pm \sqrt{3} \)
(c) \( c > 0 \)
(d) \( 0 < c < 1 \)
Answer: (b) \( c = \pm \sqrt{3} \)
In simple words: The sum of the squares of the direction cosines of any line must always be equal to 1. By applying this property to the given direction cosines, we can set up an equation to solve for the value of 'c'.

🎯 Exam Tip: A fundamental property of direction cosines (l, m, n) is \( l^2 + m^2 + n^2 = 1 \). Always use this identity when dealing with problems involving direction cosines.

Question 22. The vector equation \( \overline { r } = (\hat { i } – 2\hat { j } – \hat { k }) + t(6\hat { j } – \hat { k }) \) represents a straight line passing through the points
(a) (0, 6, -1) and (1, -2, -1)
(b) (0, 6, -1) and (-1, -4, -2)
(c) (1, -2, -1) and (1, 4, -2)
(d) (1, -2, -1) and (0, -6, 1)
Answer: (c) (1, -2, -1) and (1, 4, -2)
In simple words: A line given by \( \overline{r} = \overline{a} + t(\overline{b}-\overline{a}) \) passes through point \( \overline{a} \) and also through point \( \overline{b} \). By comparing the given equation with this form, we can find the two points the line goes through. The direction vector `(6, -1)` in the hint indicates movement along the x- and z-axes, which helps define the line's path.

🎯 Exam Tip: When given a vector equation of a line, identify the fixed point (position vector) and the direction vector to find two points on the line. The fixed point is `a`, and `a + (b-a)` gives `b`.

Question 23. If the distance of the point (1, 1, 1) from the origin is half of its distance from the plane x + y + z + k = 0, then the values of k are
(a) ±3
(b) ±6
(c) 2
(d) 3, -9
Answer: (d) 3, -9
In simple words: First, calculate how far the point (1, 1, 1) is from the origin (0, 0, 0). Then, use the formula to find the distance from (1, 1, 1) to the plane \( x + y + z + k = 0 \). The problem states that the first distance is half of the second, which allows us to solve for 'k'. Understanding the geometric meaning of these distances is key to solving the problem.

🎯 Exam Tip: Remember the distance formula from a point \( (x_1, y_1, z_1) \) to the origin is \( \sqrt{x_1^2 + y_1^2 + z_1^2} \), and to a plane \( ax+by+cz+d=0 \) is \( \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \).

Question 24. If the planes \( \overline { r } \cdot (2\hat { i } – \lambda\hat { j } + \hat { k }) = 3 \) and \( \overline { r } \cdot (4\hat { i } + \hat { j } – \mu\hat { k }) = 5 \) are parallel, then the value of \( \lambda \) and \( \mu \) are
(a) \( \frac { 1 }{ 2 }, -2 \)
(b) \( -\frac { 1 }{ 2 }, 2 \)
(c) \( -\frac { 1 }{ 2 }, -2 \)
(d) \( \frac { 1 }{ 2 }, 2 \)
Answer: (c) \( -\frac { 1 }{ 2 }, -2 \)
In simple words: When two planes are parallel, their normal vectors (the vectors that are perpendicular to the planes) must point in the same direction or opposite direction. This means one normal vector is a multiple of the other. We can compare the components of the normal vectors to find the unknown values \( \lambda \) and \( \mu \). This principle ensures that the planes never intersect.

🎯 Exam Tip: For parallel planes, their normal vectors \( \overline{n_1} \) and \( \overline{n_2} \) are proportional, meaning \( \overline{n_1} = m\overline{n_2} \) for some scalar \( m \). Equate the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) to solve for the unknowns.

Question 25. If the length of the perpendicular from the origin to the plane \( 2x + 3y + \lambda z = 1 \), \( \lambda > 0 \) is \( \frac { 1 }{ 5 } \), then the value of \( \lambda \) is
(a) \( 2\sqrt{3} \)
(b) \( 3\sqrt{2} \)
(c) 0
(d) 1
Answer: (a) \( 2\sqrt{3} \)
In simple words: We use a special formula to find the shortest distance from the origin to a flat surface (a plane). We know this distance is \( \frac{1}{5} \). By putting the numbers from the plane's equation into the formula and solving, we can find the hidden value of \( \lambda \). The condition \( \lambda > 0 \) helps confirm our final answer.

🎯 Exam Tip: The distance from the origin \( (0,0,0) \) to the plane \( Ax+By+Cz+D=0 \) is given by \( \frac{|D|}{\sqrt{A^2+B^2+C^2}} \). Make sure to rewrite the plane equation as \( 2x+3y+\lambda z - 1 = 0 \) so \( D = -1 \).