If vector of a =2i+2j+3k, b=-i+2j+k and c=3i+j are such that a+λb is perpendicular to c , then find the value of λ.
Answers
If vector of a =2i+2j+3k, b=-i+2j+k and c=3i+j then value of λ is 8
Solution:
Given, vector of a = 2i + 2j + 3k, b = -I + 2j + k and c = 3i + j such that a + λb is perpendicular to c
Then we have to find the value of λ.
Now, as a + λb is perpendicular to c, their dot product is equal to zero
→ (a + λb) . ( c) = 0
a.c + λb.c = 0
Now substitute a, b, c values in above equation.
(2i + 2j + 3k).( 3i + j) + λ( -i + 2j + k).(3i + j) = 0
On simplification we get
(6 + 2) + λ( -3 + 2) = 0 [since we know that, i.i = 1 and i.j = i.k = 0 ]
8 + λ( -1) = 0
8 – λ = 0
λ = 8
hence, the value of λ is 8.
Answer:
If vector of a =2i+2j+3k, b=-i+2j+k and c=3i+j then value of λ is 8
Solution:
Given, vector of a = 2i + 2j + 3k, b = -I + 2j + k and c = 3i + j such that a + λb is perpendicular to c
Then we have to find the value of λ.
Now, as a + λb is perpendicular to c, their dot product is equal to zero
→ (a + λb) . ( c) = 0
a.c + λb.c = 0
Now substitute a, b, c values in above equation.
(2i + 2j + 3k).( 3i + j) + λ( -i + 2j + k).(3i + j) = 0
On simplification we get
(6 + 2) + λ( -3 + 2) = 0 [since we know that, i.i = 1 and i.j = i.k = 0 ]
8 + λ( -1) = 0
8 – λ = 0
λ = 8
Step-by-step explanation: