If vector r → is perpendicular to both α → = 4 i ^ + 5 j ^ - k ^ and β → = i ^ - 4 j ^ + 5 k ^ and satisfies the condition r → · γ → = 21 , where γ → = 3 i ^ + j ^ - k ^ . Find r → .

Given vectors: α → = 4 i ^ + 5 j ^ - k ^ , β → = i ^ - 4 j ^ + 5 k ^ and γ → = 3 i ^ + j ^ - k ^ Let r → = x i ^ + y j ^ + z k ^ be the required vector. And it is given that, r → is perpendicular to the vectors α → and β → i.e., r → · α → = r → · β → = 0 Now, r → · α → = 0 ⇒ x i ^ + y j ^ + z k ^ · 4 i ^ + 5 j ^ - k ^ = 0 ⇒ 4 x + 5 y - z = 0 → 1 And r → · β → = 0 ⇒ ( x i ^ + y j ^ + z k ^ ) · ( i ^ - 4 j ^ + 5 k ^ ) = 0 ⇒ x - 4 y + 5 z = 0 → 2 Again r → · γ → = 21 ⇒ ( x i ^ + y j ^ + z k ^ ) · ( 3 i ^ + j ^ - k ^ ) = 21 ⇒ 3 x + y - z = 21 → 3 Now, equation 1 - equation 3 , then we get: x + 4 y = - 21 → 4 Again equation 2 + equation 3 × 5 , then we get: 16 x + y = 105 → 5 On solving 4 and 5 , then we get: x = 7 ; y = - 7 Substituting x = 7 ; y = - 7 in equation 3 , then we get: z = - 7 Hence, the required vector is r → = 7 ( i ^ - j ^ - k ^ ) .

Prove by vector method, the trigonometric formula: cos A = b 2 + c 2 - a 2 2 b c .

Here, | B C → | = a , | C A → | = b , | B A → | = c Now, by triangle law of vectors, from triangle A B C , we have: A B → + B C → + C A → = 0 → Therefore, A B → + B C → = A C → ⇒ B C → = A C → - A B → = B A → - C A → [Sum of vectors represented by three sides of a triangle taken in order is zero] ⇒ B C → · B C → = ( B A → - C A → ) · ( B A → - C A → ) ⇒ B C → 2 = | C A → | 2 + | B A → | 2 - 2 C A → · B A → ⇒ a → 2 = b → 2 + c → 2 - 2 b → · c → ⇒ a → 2 = b → 2 + c → 2 - 2 | b → ‖ c → | cos A ⇒ a 2 = b 2 + c 2 - 2 b c cos A ⇒ cos A = b 2 + c 2 - a 2 2 b c Hence, proved.

Prove by vector method, the trigonometric formula: cos B = a 2 + c 2 - b 2 2 a c .

Here, | B C → | = a , | C A → | = b , | A B → | = c Now, by triangle law of vectors, from triangle A B C , we have: A B → + B C → + C A → = 0 → i.e., a → + b → + c → = 0 → [Sum of vectors represented by three sides of a triangle taken in order is zero] ⇒ a → + c → = - b → Squaring both sides, then we have: | a → + c → | 2 = | - b → | 2 ⇒ | a → | 2 + | c → | 2 + 2 · a → · c → = | b → | 2 ⇒ | a → | 2 + | c → | 2 + 2 | a → | · | c → | cos ( π - B ) = | b → | 2 ⇒ | a → | 2 + | c → | 2 - 2 | a → | · | c → | · cos B = | b → | 2 ⇒ a 2 + c 2 - 2 a c cos B = b 2 [Since | a → | = a , | b → | = b , | c → | = c ] ⇒ cos B = a 2 + c 2 - b 2 2 a · c Hence, proved.

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Vectors>Chapter 23 - Vectors>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 2: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 2: EXERCISE

Attempt the practice questions from Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 2: EXERCISE with Hints & Solutions

EASY

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 23 - Vectors>EXERCISE>Q 3

The three vectors $7 \hat{i} - 11 \hat{j} + \hat{k}$ , $5 \hat{i} + 3 \hat{j} - 2 \hat{k}$ and $12 \hat{i} - 8 \hat{j} - \hat{k}$

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 23 - Vectors>EXERCISE>Q 16

If vector $\vec{r}$ is perpendicular to both $\vec{α} = 4 \hat{i} + 5 \hat{j} - \hat{k}$ and $\vec{β} = \hat{i} - 4 \hat{j} + 5 \hat{k}$ and satisfies the condition $\vec{r} \cdot \vec{γ} = 21$ , where $\vec{γ} = 3 \hat{i} + \hat{j} - \hat{k}$ . Find $\vec{r}$ .