Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 6 Line and Plane Miscellaneous Exercise 6A Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6A
Question 1
Maharashtra Board Solution
Find the vector equation of the line passing through the point having position vector and parallel to .
Solution & Step-by-Step Answer:
The vector equation of the line passing through A() and parallel to the vector is , where λ is a scalar. ∴ the vector equation of the line passing through the point having position vector and parallel to the vector is .
Question 2
Maharashtra Board Solution
Find the vector equation of the line which passes through the point (3, 2, 1) and is parallel to the vector .
Solution & Step-by-Step Answer:
The vector equation of the line passing through A() and parallel to the vector is , where λ is a scalar. ∴ the vector equation of the line passing through the point having position vector and parallel to the vector
Question 3
Maharashtra Board Solution
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and parallel to the line
Solution & Step-by-Step Answer:
The line has direction ratios 3, 5, 6. The required line has direction ratios 3, 5, 6 as it is parallel to the given line. It passes through the point (-2, 4, -5). The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are
Question 4
Maharashtra Board Solution
Obtain the vector equation of the line .
Solution & Step-by-Step Answer:
The cartesian equations of the line are . This line is passing through the point A(-5, -4, -5) and having direction ratios 3, 5, 6. Let be the position vector of the point A w.r.t. the origin and be the vector parallel to the line. Then and . The vector equation of the line passing through A() and parallel to is where λ is a scalar. ∴ the vector equation of the required line is
Question 5
Maharashtra Board Solution
Find the vector equation of the line which passes through the origin and the point (5, -2, 3).
Solution & Step-by-Step Answer:
Let be the position vector of the point B(5, -2, 3). Then Origin has position vector . The vector equation the line passing through A() and B() is where λ is a scalar. ∴ the vector equation of the required line is
Question 6
Maharashtra Board Solution
Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6).
Solution & Step-by-Step Answer:
Let A = (3, -2, -5), B = (3, -2, 6) The direction ratios of the line AB are 3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11. The parametric equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are x = x1 + aλ, y = y1 + bλ, z = z1 + cλ ∴ the parametric equattions of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ i.e. x = 3, y = -2, z = 11λ – 5 ∴ the cartesian equations of the line are x = 3, y = -2, z = 11λ – 5, λ is a scalar.
Question 7
Maharashtra Board Solution
Find the Cartesian equations of the line passing through A(3, 2, 1) and B(1, 3, 1).
Solution & Step-by-Step Answer:
The direction ratios of the line AB are 3 – 1, 2 – 3, 1 – 1 i.e. 2, -1, 0. The parametric equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are x = x1 + aλ, y = y1 + bλ, z = z1 + cλ ∴ the parametric equattions of the line passing through (3, 2, 1) and having direction ratios 2, -1, 0 are x = 3 + 2λ, y = 2 – λ, z = 1 + 0(λ) x – 3 = 2λ, y – 2 = -λ, z = 1 ∴ = λ, z = 1 ∴ the cartesian equations of the line are , z = 1.
Question 8
Maharashtra Board Solution
Find the Cartesian equations of the line passing through the point A(1, 1, 2) and perpendicular to vectors and .
Solution & Step-by-Step Answer:
Let the required line have direction ratios p, q, r., It is perpendicular to the vectors and . ∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1. ∴ p + 2q + r = 0, 3p + 2q – r = 0 ∴ the required line has direction ratios -1, 1, -1. The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are ∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are
Question 9
Maharashtra Board Solution
Find the Cartesian equations of the line which passes through the point (2, 1, 3) and perpendicular to lines and .
Solution & Step-by-Step Answer:
Let the required line have direction ratios p, q, r. It is perpendicular to the vector and . ∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1. ∴ p + 2q + r = 0, 3 + 2q – r = 0 ∴ the required line has direction ratios 2, -7, 4. The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are ∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are
Question 10
Maharashtra Board Solution
Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.
Solution & Step-by-Step Answer:
The given line is = λ … (Say) ∴ coordinates of any point on the line are x = λ + 1, y = λ + 2, z = λ + 3 ∴ position vector of any point on the line is (λ + 1) + (λ + 2) + (λ + 3) … (1) If is parallel to the given line whose direction ratios are 1, 1, 1, then . Let the required line passing through O meet the given line at M. ∴ position vector of M = = (λ + 1) + (λ + 2) + (λ + 3) … [By (1)] The required line is perpendicular to given line The vector equation of the line passing through A() and B() is , λ is a scalar. ∴ the vector equation of the line passing through o() and M() is where λ is a scalar. Hence, vector equation of the required line is .
Question 11
Maharashtra Board Solution
Find the value of λ so that lines and are at right angle.
Solution & Step-by-Step Answer:
The equations of the given lines are
Question 12
Maharashtra Board Solution
Find the acute angle between lines and .
Solution & Step-by-Step Answer:
Question 13
Maharashtra Board Solution
Find the acute angle between lines x = y, z = 0 and x = 0, z = 0.
Solution & Step-by-Step Answer:
The equations x = y, z = 0 can be written as , z = 0 ∴ the direction ratios of the line are 1, 1, 0. The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0. ∴ its directiton ratios are 0, 1, 0. Let and be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0. If θ is the acute angle between the lines, then
Question 14
Maharashtra Board Solution
Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0.
Solution & Step-by-Step Answer:
The equations x = -y, z = 0 can be written as , z = 0. ∴ the direction ratios of the line are 1, 1, 0. The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0. ∴ its direction ratios are 0, 1, 0. Let and be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0
Question 15
Maharashtra Board Solution
Find the co-ordinates of the foot of the perpendicular drawn from the point (0, 2, 3) to the line .
Solution & Step-by-Step Answer:
Let P = (0, 2, 3) Let M be the foot of the perpendicular drawn from P to the line = λ ……(Say) The coordinates of any point on the line are given by x = 5λ – 3, y = 2λ + 1, z = 3λ – 4 Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1) The direction ratios of PM are 5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7 Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3, 5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0 25λ – 15 + 4λ – 2 + 9λ – 21 =0 38λ – 38 = 0 ∴ λ = 1 Substituting λ = 1 in (1), we get. M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1). Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).
Question 16
Maharashtra Board Solution
By computing the shortest distance determine whether following lines intersect each other. (i) and
Solution & Step-by-Step Answer:
The shortest distance between the lines Shortest distance between the lines is 0. ∴ the lines intersect each other.
(ii) and x – 6 = y – 8 = z + 2.
Solution:
The shortest distance between the lines
∴ x1= 5, y1= 7, z1= 3, x2= 6, y2= 8, z2= 2,
l1= 4, m1= 5, n1= 1, l2= 1, m2= -2, n2= 1
= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64
= -116
and
(m1n2– m2n1)2+ (l2n1– l1n2)2+ (l1m2– l2m1)2
= (-6 + 2)2+ (1 – 7)2+ (1 – 7)2+ (-14 + 6)2
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines
or
Shortest distance between the lines is 0.
∴ the lines intersect each other.
Question 17
Maharashtra Board Solution
If lines and intersect each other then find m.
Solution & Step-by-Step Answer:
Here, (x1, y1, z1) ≡ (1, -1, 1), (x2, y2, z2) ≡ (2, -m, 2), a1 = 2, b1 = 3, c1 = 4, a2 = 1, b2 = 2, c2 = 1 Substituting these values in (1), we get ∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0 ∴ -5 + 2 – 2m + 1 = 0 ∴ -2m = 2 ∴ m = -1.
Question 18
Maharashtra Board Solution
Find the vector and Cartesian equations of the line passing through the point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2.
Solution & Step-by-Step Answer:
Let be the position vector of the point A (-1, -1, 2) w.r.t. the origin. Then The equation of given line is x – 2 = 3y + 1 = 6z – 2. The direction ratios of this line are
Question 19
Maharashtra Board Solution
Find the direction cosines of the line .
Solution & Step-by-Step Answer:
Question 20
Maharashtra Board Solution
Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersects the .
Solution & Step-by-Step Answer:
Let the required line have direction ratios a, b, c Since the line passes through the origin, its cartesian equations are …(1) This line is perpendicular to the line x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1. ∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0 ∴ 4b – 3c + 4a – 2c + 3a – 2b = 0 ∴ 7a + 2b – 5c = 0 From (2) and (3), we get
Question 21
Maharashtra Board Solution
Write the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.
Solution & Step-by-Step Answer:
4x – 3z + 5 = 0 can be written as This line passes through the point A(0, 2, ) position vector is Also the line has direction ratio 3, 0, 4. If is a vector parallel to the line, then The vector equation of the line passing through A() and parallel to is where λ is scalar, ∴ the vector equation of the required line is .
Question 22
Maharashtra Board Solution
Find the co-ordinates of points on the line which are at the distance 3 unit from the base point A(1, 2, 3).
Solution & Step-by-Step Answer:
The cartesian equations of the line are = λ The coordinates of any point on this line are given by x = λ + 1, y = -2λ + 2, z = 2λ + 3 Let M(λ + 1, -2λ + 2, 2λ + 3) … (1) be the point on the line whose distance from A(1, 2, 3) is 3 units. When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)] i. e. M = (2, 0, 5) When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)] i. e. M = (0, 4, 1) Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).
