Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 8 Vector Algebra – I Ex 8.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.3

Question 1.

Find \(\vec{a}\) . \(\vec{b}\) when

(i) \(\vec{a}\) = î – 2ĵ + k̂ and \(\vec{b}\) = 3î – 4ĵ – 2k̂

(ii) \(\vec{a}\) = 2î + 2ĵ – k̂ and \(\vec{b}\) = 6î – 3ĵ + 2k̂

Answer:

(i) \(\vec{a}\) = î – 2ĵ + k̂ and \(\vec{b}\) = 3î – 4ĵ – 2k̂

\(\vec{a}\) . \(\vec{b}\) = (î – 2ĵ + k̂) . (3î – 4ĵ – 2k̂)

= (1) (3) + (-2) (-4) + (1) (-2)

= 3 + 8 – 2

= 9

(ii) \(\vec{a}\) = 2î + 2ĵ – k̂ and \(\vec{b}\) = 6î – 3ĵ + 2k̂

\(\vec{a}\) . \(\vec{b}\) = (2î + 2ĵ – k̂) . (6î – 3ĵ + 2k̂)

= (2) (6) + (2) (-3) + (-1) (2)

= 12 – 6 – 2 = 12 – 8 = 4

Question 2.

Find the value λ for which the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular, where

(i) \(\vec{a}\) = 2î + λĵ – k̂ and \(\vec{b}\) = î – 2ĵ + 3k̂

(ii) \(\vec{a}\) = 2î + 4ĵ – k̂ and \(\vec{b}\) = 3î – 2ĵ + λk̂

Answer:

When \(\vec{a}\) and \(\vec{b}\) are ⊥^{r} then \(\vec{a} \cdot \vec{b}\) = 0

\(\vec{a}\) ⊥^{r} \(\vec{b}\) ⇒ \(\vec{a} \cdot \vec{b}\) = 0

(i) (2) (1) + (λ) (-2) + (1) (3) = 0 ⇒ λ = 5/2

(ii) (2) (3) + (4) (-2) + (-1) (λ) = 0

6 – 8 – λ = 0

-λ – 2 = 0 ⇒ -λ = 2 ⇒ λ = -2

Question 3.

Answer:

Question 4.

Find the angle between the vectors

(i) 2î + 3ĵ – 6k̂ and 6î – 3ĵ + 2k̂

(ii) î – ĵ and ĵ – k̂

Answer:

(i) 2î + 3ĵ – 6k̂ and 6î – 3ĵ + 2k̂

Let θ be the angle between the given vectors, then

(ii) î – ĵ and ĵ – k̂

Let θ be the angle between the given vectors, then

Question 5.

Answer:

Question 6.

Show that the vectors \(\vec{a}\) = 2î + 3ĵ + 6k̂ \(\vec{b}\) = 6î + 2ĵ – 3k̂ and \(\vec{c}\) = 3î – 6ĵ + 6k̂ are mutually orthogonal.

Answer:

Given \(\vec{a}\) = 2î + 3ĵ + 6k̂ \(\vec{b}\) = 6î + 2ĵ – 3k̂ and \(\vec{c}\) = 3î – 6ĵ + 6k̂

\(\vec{a}\) . \(\vec{b}\) = (2î + 3ĵ + 6k̂) . (6î + 2ĵ – 3k̂)

= (2) (6) + (3) (2) + (6) (-3)

= 12 + 6 – 18

= 0

∴ \(\vec{a}\) and \(\vec{a}\) are perpendicular.

\(\vec{b}\) . \(\vec{c}\) = (6î + 2ĵ – 3k̂) . (3î – 6ĵ + 6k̂)

= (6) (3) + (2) (- 6) + (-3) (2)

= 18 – 12 – 6

= 0

∴ \(\vec{b}\) and \(\vec{c}\) are perpendicular.

\(\vec{c}\) . \(\vec{a}\) = (3î – 6ĵ + 6k̂) . (2î + 3ĵ + 6k̂)

= (3) (2) + (-6) (3) + (2) (6)

= 6 – 18 + 12

= 0

∴ \(\vec{c}\) and \(\vec{a}\) are perpendicular.

Hence \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors.

Question 7.

Show that the vectors – î – 2ĵ – 6k̂, 2î – ĵ + k̂ and – î + 3ĵ + 5k̂ from a right-angled triangle.

Answer:

CA = \(\sqrt{1+9+25}\) = \(\sqrt{35}\)

AB ≠ BC + CA

∴ The given vectors form a triangle, Also

AB^{2} = 41, BC^{2} = 6, CA^{2} = 35

AB^{2} = BC^{2} + CA^{2}

∴ ∆ ABC is a right angled triangle.

Question 8.

Answer:

Question 9.

Show that the points (2, – 1, 3), (4, 3, 1), and (3, 1, 2) are collinear.

Answer:

Let the given points be

A (2, -1, 3), B (4, 3, 1) and C (3, 1, 2)

AB = 2√6, BC = √6, CA = √6

BC + CA = √6 + √6 = 2√6

∴ BC + CA = BA = 2√6

Hence the given points A, B, C are collinear.

Question 10.

If \(\vec{a}\), \(\vec{b}\) are unit vectors and θ is the angle between them, show that

(i)

(ii)

(iii)

Answer:

Given \(\vec{a}\) and \(\vec{b}\) are unit vectors.

∴ |\(\vec{a}\)| = 1 and |\(\vec{b}\)| = 1

(i)

(ii)

(iii)

Question 11.

Answer:

Question 12.

Find the projection of the vector î + 3ĵ + 7k̂ on the vector 2î + 6ĵ + 3k̂

Answer:

The given vectors are î + 3ĵ + 7k̂ and 2î + 6ĵ + 3k̂

Projection of î + 3ĵ + 7k̂ on 2î + 6ĵ + 3k̂ is

Question 13.

Find λ, when the projection \(\vec{a}\) = λî + ĵ + 4k̂ on \(\vec{b}\) = 2î + 6ĵ + 3k̂ is 4 units.

Answer:

The given vectors are

\(\vec{a}\) = λî + ĵ + 4k̂ , \(\vec{b}\) = 2î + 6ĵ + 3k̂

Also given that projection of \(\vec{a}\) and \(\vec{b}\) is 4 units.

Question 14.

Answer: