Question

Let S be the set of all (λ, μ) for which the vectors $ λ {i}ˆ-jˆ+kˆ, iˆ +2jˆ+µkˆ and 3iˆ -4jˆ +5kˆ, where λ-μ = 5, are coplanar, then $$ \sum_{(λ, μ) εs}80(λ^2, μ^2) $ is equal to

• A. 2130
• B. 2210
• C. 2290
• D. 2370

Answer 1

The Correct Option is C

To solve this problem, we need to determine the conditions under which the given vectors are coplanar. Three vectors are coplanar if their scalar triple product is zero. Let's analyze the vectors provided:

• \(\mathbf{a} = \lambda \hat{i} - \hat{j} + \hat{k}\)
• \(\mathbf{b} = \hat{i} + 2\hat{j} + \mu \hat{k}\)
• \(\mathbf{c} = 3\hat{i} - 4\hat{j} + 5\hat{k}\) 

The scalar triple product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0\) for coplanarity.

Calculate the cross product \(\mathbf{b} \times \mathbf{c}\):

The determinant of the following matrix gives the cross-product:

\hat{i}	\hat{j}	\hat{k}
1	2	\mu
3	-4	5

Calculate:

• Coefficient of \(\hat{i}\): \(10 + 4\mu\)
• Coefficient of \(\hat{j}\): \(-(1 \cdot 5 - \mu \cdot 3) = -(5 - 3\mu) = 3\mu - 5\)
• Coefficient of \(\hat{k}\): \((-4 - 6) = -10\)

Thus, \(\mathbf{b} \times \mathbf{c} = (10 + 4\mu)\hat{i} + (3\mu - 5)\hat{j} - 10\hat{k}\).

Now calculate the dot product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\):

\(\lambda(10 + 4\mu) - 1(3\mu - 5) + 1(-10) = 0\)

Solving, we get:

\(\lambda(10 + 4\mu) - 3\mu + 5 - 10 = 0\)

\(\lambda(10 + 4\mu) - 3\mu - 5 = 0\)

Substituting \(\lambda = \mu + 5\), we have:

\({(\mu + 5)(10 + 4\mu) - 3\mu - 5 = 0}\)

After simplification:

\({10\mu + 4\mu^2 + 50 + 20\mu - 3\mu - 5 = 0}\)

\({4\mu^2 + 27\mu + 45 = 0}\)

Solve the quadratic equation:

Using the quadratic formula \(\mu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\({b = 27, a = 4, c = 45}\)

\(\mu = \frac{-27 \pm \sqrt{27^2 - 4 \cdot 4 \cdot 45}}{8}\)

Calculate:

\(729 - 720 = 9\)

\(\mu = \frac{-27 \pm 3}{8}\)

Thus:

• \(\mu = \frac{-27 + 3}{8} = \frac{-24}{8} = -3\)
• \(\mu = \frac{-27 - 3}{8} = \frac{-30}{8} = -\frac{15}{4}\)

Therefore, we have two solutions for \(\mu\): \(-3\) and \(-\frac{15}{4}\).

For each \(\mu\):

• When \(\mu = -3\), \(\lambda = 2\).
• When \(\mu = -\frac{15}{4}\), \(\lambda = \frac{5}{4}\).

The solutions yield:

\(\sum_{(\lambda, \mu) \, \varepsilon \, S} 80(\lambda^2 + \mu^2)\)

• For \((\lambda, \mu) = (2, -3)\), \(80((2)^2 + (-3)^2) = 80(4 + 9) = 80 \times 13 = 1040\)
• For \((\lambda, \mu) = \left( \frac{5}{4}, -\frac{15}{4} \right)\), \(80 \left( \left( \frac{5}{4} \right)^2 + \left( -\frac{15}{4} \right)^2 \right) = 80 \left( \frac{25}{16} + \frac{225}{16} \right) = 80 \times \frac{250}{16} = 80 \times \frac{125}{8} = 1250\)

Therefore, the sum is:

\(1040 + 1250 = 2290\)

Hence, the answer is \(2290\).