Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5

Question 1.
Find the parametric form of vector equation and Cartesian equations of straight line passing through (5, 2, 8) and is perpendicular to the straight lines
\(\overline { r }\) = (\(\hat { i }\) + \(\hat { j }\) – \(\hat { k }\)) + s(2\(\hat { i }\) – 2\(\hat { j }\) + \(\hat { k }\)) and
\(\overline { r }\) = (2\(\hat { i }\) – \(\hat { j }\) – 3\(\hat { k }\)) + t(\(\hat { i }\) + 2\(\hat { j }\) + 2\(\hat { k }\)).
Solution:
Given points (5, 2, 8)
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 1
2\(\hat { i }\) + \(\hat { j }\) + 3\(\hat { k }\) is a vector perpendicular to both the given straight lines it passes through (5, 2, 8)
The equation is \(\overline { r }\) = (5\(\hat { i }\) + 2\(\hat { j }\) + 8\(\hat { k }\)) + t(2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\))
Cartesian form = \(\frac { x-5 }{ 2 }\) = \(\frac { y-2 }{ 1 }\) = \(\frac { z-8 }{ -2 }\)

Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5

Question 2.
Show that the lines
\(\overline { r }\) = (6\(\hat { i }\) + \(\hat { j }\) + 2\(\hat { k }\)) + s(\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)) and
\(\overline { r }\) = (3\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)) + t(2\(\hat { i }\) + 4\(\hat { j }\) – 5\(\hat { k }\)) are skew lines and hence find the shortest distance between them.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 2
\(\overline { b }\) is not a scalar multiple of \(\overline { d }\)
∴ They are not parallel.
∴ The given lines are skew lines.
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 3

Question 3.
If the two lines \(\frac { x-1 }{ 2 }\) = \(\frac { y+1 }{ 3 }\) = \(\frac { z-1 }{ 4 }\) and \(\frac { x-3 }{ 1 }\) = \(\frac { y-m }{ 2 }\) = z intersect at a point, find the value of m.
Solution:
(x1, y1, z1) = (1, -1, 1), (x2, y2, z2) = (3, m, 0)
(1, 2, b3) = (2, 3, 4), (d1, d2, d3) = (1, 2, 1).
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 4
2(3 – 8) – (m + 1) (2 – 4) – 1(4 – 3) = 0
-10 – (m + 1) (-2) – 1(1) = 0
-10 + 2m + 2 – 1 = 0
2m – 9 = 0
2m = 9
m = \(\frac { 9 }{ 2 }\)

Question 4.
Show that the lines \(\frac { x-3 }{ 3 }\) = \(\frac { y-3 }{ -1 }\), z – 1 = 0 and \(\frac { x-6 }{ 2 }\) = \(\frac { z-1 }{ 3 }\), y – 2 = 0 intersect. Also find the point of intersection.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 5
(2s + 6, 2, 3s + 1) ……… (2)
(1) = (2)
((3t + 3), -t + 3, 1) = (2s + 6, 2, 3s + 1)
Compare on both sides
-t + 3 = 2; 3s + 1 = 1
t = 3 – 2; 3s = 0
t = 1; s = 0
(1) ⇒ Point of intersect (6, 2, 1)

Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5

Question 5.
Show that the straight lines x + 1 = 2y = -12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them.
Solution:
Given x + 1 = 2y = -12z
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 6
\(\overline { b }\) is not a scalar multiple of \(\overline { d }\). So, the two vectors are not parallel.
∴ The given lines are skew lines.
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 7

Question 6.
Find the parametric form of vector equation of the straight line passing through (-1, 2, 1) and parallel to the straight line
\(\overline { r }\) = (2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\)) + t(\(\hat { i }\) – 2\(\hat { j }\) + \(\hat { k }\)) and hence find the shortest distance between the lines.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 8

Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5

Question 7.
Find the foot of the perpendicular drawn: from the point (5, 4, 2) to the line \(\frac { x+1 }{ 2 }\) = \(\frac { y-3 }{ 3 }\) = \(\frac { z-1 }{ -1 }\). Also, find the equation of the perpendicular.
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 9
Solution:
\(\overline { r }\) = \(\overline { a }\) + t\(\overline { b }\)
\(\overline { a }\) = -i + 3j + k, \(\overline { b }\) = 2i + 3j – k
Given points D (5, 4, 2) to the point A. If P is the foot of the perpendicular from to the straight line.
F is of the form
(2t – 1, 3t + 3, -t + 1) and
\(\overline { DF }\) = \(\overline { OF }\) – \(\overline { OD }\) = (2t – 6)i + (3t – 1)j + (-t – l)k
\(\overline { b }\) is perpendicular to \(\overline { DF }\), we have
\(\overline { b }\).\(\overline { DF }\) = 0 => (2t – 6) 2 + 3(3t – 1) – 1(-t – 1) = 0
4t- 12 + 9t – 3 + t + 1 = 0
14t – 14 = 0
14t = 14
t = 1
∴ F (2 – 1, 3 + 3, -1 + 1) = F (1, 6, 0) is foot point. Equation of the perpendicular.
(x1, y1, z1) = (5, 4, 2), (x2, y2, z2) = (1, 6, 0).
Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5 10

Samacheer Kalvi 12th Maths Guide Chapter 6 Applications of Vector Algebra Ex 6.5