The vectors a, b, and c are mutually orthogonal if their dot products are zero. This means that:

a.b = 0
a.c = 0
b.c = 0

Let's calculate these dot products:

1. a.b = (iˆ−jˆ+2kˆ).(2iˆ+4jˆ+kˆ) = 2*1 - 1*4 + 2*1 = 0
2. a.c = (iˆ−jˆ+2kˆ).(λiˆ+jˆ+μkˆ) = λ*1 - 1*1 + 2*μ = 0
3. b.c = (2iˆ+4jˆ+kˆ).(λiˆ+jˆ+μkˆ) = 2*λ + 4*1 + μ*1 = 0

From the second equation, we get λ = 1 - 2μ. Substituting this into the third equation, we get:

2*(1 - 2μ) + 4 + μ = 0
2 - 4μ + 4 + μ = 0
-3μ + 6 = 0
μ = 2

Substituting μ = 2 into the equation λ = 1 - 2μ, we get λ = 1 - 2*2 = -3.

So, (λ,μ) = (-3,2), which corresponds to the fourth option.

Question

The vectors a, b, and c are mutually orthogonal if their dot products are zero. This means that:

a.b = 0
a.c = 0
b.c = 0

Let's calculate these dot products:

1. a.b = (iˆ−jˆ+2kˆ).(2iˆ+4jˆ+kˆ) = 2*1 - 1*4 + 2*1 = 0
2. a.c = (iˆ−jˆ+2kˆ).(λiˆ+jˆ+μkˆ) = λ*1 - 1*1 + 2*μ = 0
3. b.c = (2iˆ+4jˆ+kˆ).(λiˆ+jˆ+μkˆ) = 2*λ + 4*1 + μ*1 = 0

From the second equation, we get λ = 1 - 2μ. Substituting this into the third equation, we get:

2*(1 - 2μ) + 4 + μ = 0
2 - 4μ + 4 + μ = 0
-3μ + 6 = 0
μ = 2

Substituting μ = 2 into the equation λ = 1 - 2μ, we get λ = 1 - 2*2 = -3.

So, (λ,μ) = (-3,2), which corresponds to the fourth option.

Knowee AI · Accepted Answer

The vectors a, b, and c are mutually orthogonal if their dot products are zero. This means that:

a.b = 0
a.c = 0
b.c = 0

Let's calculate these dot products:

1. a.b = (iˆ−jˆ+2kˆ).(2iˆ+4jˆ+kˆ) = 2*1 - 1*4 + 2*1 = 0
2. a.c = (iˆ−jˆ+2kˆ).(λiˆ+jˆ+μkˆ) = λ*1 - 1*1 + 2*μ = 0
3. b.c = (2iˆ+4jˆ+kˆ).(λiˆ+jˆ+μkˆ) = 2*λ + 4*1 + μ*1 = 0

From the second equation, we get λ = 1 - 2μ. Substituting this into the third equation, we get:

2*(1 - 2μ) + 4 + μ = 0
2 - 4μ + 4 + μ = 0
-3μ + 6 = 0
μ = 2

Substituting μ = 2 into the equation λ = 1 - 2μ, we get λ = 1 - 2*2 = -3.

So, (λ,μ) = (-3,2), which corresponds to the fourth option.

If the vectors a→=iˆ−jˆ+2kˆ,b→=2iˆ+4jˆ+kˆ and c→=λiˆ+jˆ+μkˆ are mutually orthogonal, then (λ,μ)=(2,−3)(−2,3)(3,−2)(−3,2)

Question

If the vectors $\mathbf{a} = \hat{i} - \hat{j} + 2\hat{k}, \mathbf{b} = 2\hat{i} + 4\hat{j} + \hat{k}$ and $\mathbf{c} = \lambda \hat{i} + \hat{j} + \mu \hat{k}$ are mutually orthogonal, then $(\lambda, \mu) = (2, -3)(-2, 3)(3, -2)(-3, 2)$

Solution

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