Question

5. If Ā= 2i +3j+6k and B = 3i– 6j + 2k then vector
perpendicular to both A and B has magnitude k times
of (6î +2j–3k). That k is equal to
(a) 1
(b)4
(c)7
(d) 9​

Answer 1

answer : option (c) 7

explanation : it is given that , A = 2i + 3j + 6k and B = 3i - 6j + 2k.

Let a vector R which is perpendicular to both A and B. it means, R is the cross product of A and B.

i.e., R = ± (A × B)

or, |R| = |(A × B)|

so, magnitude of cross product of A and B, |(A × B)| = |A| |B| sinα , where α is angle between A and B.

|A| = √(2² + 3² + 6²) = √(4 + 9 + 36) = 7

|B| = √{3² + (-6)² + 2²} = 7

and angle between them, α = cos^-1$\left(\frac{A.B}{|A||B|}\right)$

= cos^-1{(2i + 3j + 6k).(3i - 6j + 2k)}/7.7}

= cos^-1{(6 - 18 + 12)/7.7}

= cos^-1(0) = π/2

so, angle between them, α = 90°

now, |R| = |(A × B)| = |A||B|sinα

= 7.7 sin90° = 49 unit.

a/c to question,

|R| = k|(6i + 2j -3k)|

or, 49 = k√{(-6)² + (2)² + (-3)²}

or, 49 = k × 7

or, k = 7

hence, value of k = 7

Answer 2

Answer:

Explanation:

answer : option (c) 7

explanation : it is given that , A = 2i + 3j + 6k and B = 3i - 6j + 2k.

Let a vector R which is perpendicular to both A and B. it means, R is the cross product of A and B.

i.e., R = ± (A × B)

or, |R| = |(A × B)|

so, magnitude of cross product of A and B, |(A × B)| = |A| |B| sinα , where α is angle between A and B.

|A| = √(2² + 3² + 6²) = √(4 + 9 + 36) = 7

|B| = √{3² + (-6)² + 2²} = 7

and angle between them, α = cos^-1

= cos^-1{(2i + 3j + 6k).(3i - 6j + 2k)}/7.7}

= cos^-1{(6 - 18 + 12)/7.7}

= cos^-1(0) = π/2

so, angle between them, α = 90°

now, |R| = |(A × B)| = |A||B|sinα

= 7.7 sin90° = 49 unit.

a/c to question,

|R| = k|(6i + 2j -3k)|

or, 49 = k√{(-6)² + (2)² + (-3)²}

or, 49 = k × 7

or, k = 7

hence, value of k = 7