Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.8 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8

Question 1.

Show that the straight lines

\(\overline { r }\) = (5\(\hat { i }\) + 7\(\hat { j }\) – 3\(\hat { k }\)) + s(4\(\hat { i }\) + 4\(\hat { j }\) – 5\(\hat { k }\)) and

\(\overline { r }\) = (8\(\hat { i }\) + 4\(\hat { j }\) + 5\(\hat { k }\)) + t(7\(\hat { i }\) + \(\hat { j }\) + 3\(\hat { k }\)) are coplanar. Find the vector equation of the, plane in which they lie.

Solution:

Question 2.

Show that the lines \(\frac { x-2 }{ 1 }\) = \(\frac { y-3 }{ 1 }\) = \(\frac { z -4}{ 3 }\) and \(\frac { x-1 }{ -3 }\) = \(\frac { y-4 }{ 2 }\) = \(\frac { z-5 }{ 1 }\) are coplanar. Also, find the plane containing these lines.

Solution:

Cartesian equation

x + 2y – 2z = 4

x + 2y – 2z – 4 = 0

Question 3.

If the straight lines \(\frac { x-1 }{ 1 }\) = \(\frac { y-2 }{ 2 }\) = \(\frac { z-3}{ m^2 }\) and \(\frac { x-3 }{ 1 }\) = \(\frac { y-2 }{ m^2 }\) = \(\frac { z-1 }{ 2 }\) are coplanar, find the distinct real values of m

Solution:

2(4 – m^{4}) – 2(m² – 2) = 0

8 – 2m^{4} – 2m² + 4 = 0

12 – 2m^{4} – 2m² = 0

(÷ -2) -6 + m^{4} + m² = 0

m^{4} + m² – 6 = 0

(m² – 2)(m² + 3) = 0

m² – 2 = 2; m² = -3 (not possible)

m² = 2

m = ±√2

Question 4.

If the straight lines \(\frac { x-1 }{ 2 }\) = \(\frac { y+1 }{ λ }\) = \(\frac { z }{ 2 }\) and \(\frac { x+1 }{ 5 }\) = \(\frac { y+1 }{ 2 }\) = \(\frac { z }{ λ }\) are coplanar, find λ and equations of the planes containing these two lines.

Solution:

If the two lines are coplanar

When λ = 2

(x_{1}, y_{1}, z_{1}) = (1, -1, 0)

(b_{1}, b_{2}, b_{3}) = (2, 2, 2)

(d_{1}, d_{2}, d_{3}) = (5, 2, 2)

\(\left|\begin{array}{ccc}

x-x_{1} & y-y_{1} & z-z_{1} \\

b_{1} & b_{2} & b_{3} \\

d_{1} & d_{2} & d_{3}

\end{array}\right|\) = 0

⇒ \(\left|\begin{array}{ccc}

x-1 & y+1 & z-0 \\

2 & 2 & 2 \\

5 & 2 & 2

\end{array}\right|\) = 0

⇒ (x – 1)(0) – (y + 1)(-6) + z(6) = 0

⇒6(y + 1) – 6z = 0

⇒ 6y + 6 – 6z = 0

⇒ y – z + 1 = 0

When λ = 2

(b_{1}, b_{2}, b_{3}) = (2, -2, 2)

(d_{1}, d_{2}, d_{3}) = (5, 2, -2)

⇒ \(\left|\begin{array}{ccc}

x-1 & y+1 & z-0 \\

2 & -2 & 2 \\

5 & 2 & -2

\end{array}\right|\) = 0

⇒ (x – 1)(0) – (y + 1)(-14) + z(4 + 10) = 0

⇒ 14(y + 1) + 14z = 0

⇒ 14y + 14 + 14z = 0

⇒ y + z + 1 = 0