Question

show the vector 4i^+13j^-18k^,i^-2j^+3k^and 2i^+3j^-4k^are coplaner​

Answer 1

Answer:

Solution:If Vector A=4i+13j-18k is a linear combination of the vectors, B= i-2j+3k & C=2i+3j-4k , thenA=u B + m C4 i+13 j-18 k= u( i ...

please follow me

Answer 2

We have to show that the vectors 4i + 13j - 18k , i - 2j + 3k and 2i + 3j - 4k are coplanar.

What is coplanar ?

Coplanar vectors are the vectors which lie one the same plane.

Condition of coplanar :

• if three vectors A , B and C are coplanar vectors , then the their scalar triple product must be zero. i.e., A.(B × C) = 0
• if there are many vectors are in 3d space linearly independent then these are coplanar vectors.

here A = 4i + 13j - 18k

B = i - 2j + 3k

C = 2i + 3j - 4k

∴ (B × C) = (i - 2j + 3k) × (2i + 3j - 4k)

= i (-2 × -4 - 3 × 3) - j(1 × -4 - 3 × 2) + k(1 × 3 - 2 × -2)

= i( 8 - 9) - j(-4 - 6) + k(3 + 4)

= -i + 10j + 7k

now A.(B × C) = (4i + 13j - 18k).(-i + 10j + 7j)

= 4 × -1 + 13 × 10 - 18 × 7

= -4 + 130 - 126

= 130 - 130

= 0

∴ A.(B × C) = 0

here it is clear that, it follows the condition of coplanar , hence these all three vectors are coplanar.