To solve this problem, we will apply concepts involving the scalar triple product and vector dot product.
Given vectors:
Condition 1: \([\vec{a}\ \vec{b}\ \vec{c}] = 0\)
The scalar triple product \([\vec{a}\ \vec{b}\ \vec{c}]\) is the determinant of the matrix formed by these vectors:
| \(2\) | \(-1\) | \(1\) |
| \(1\) | \(2\) | \(-1\) |
| \(\lambda\) | \(\mu\) | \(3\) |
The determinant of this matrix is calculated as:
\(\text{Det} = 2(2 \cdot 3 + 1 \cdot \mu) - (-1)(1 \cdot 3 + \lambda \cdot -1) + 1(1 \cdot \mu - 2 \cdot \lambda)\)
\(= 2(6 + \mu) + 3 + \lambda + \mu - 2\lambda\)
\(= 12 + 2\mu + 3 + \lambda + \mu - 2\lambda\)
\(= 15 + 3\mu - \lambda\)
Since this determinant is 0, we have:
\(15 + 3\mu - \lambda = 0\)
Therefore, rearranging gives us:
\(\lambda = 3\mu + 15\) ... (Equation 1)
Condition 2: \(\vec{c} \cdot (\vec{a} + \vec{b}) = 10\)
Now compute \(\vec{a} + \vec{b}:\)
\(\vec{a} + \vec{b} = (2 + 1)\hat{i} + (-1 + 2)\hat{j} + (1 - 1)\hat{k}\)
\(= 3\hat{i} + \hat{j} + 0\hat{k}\)
The dot product \(\vec{c} \cdot (\vec{a} + \vec{b})\) is:
\((\lambda\hat{i} + \mu\hat{j} + 3\hat{k}) \cdot (3\hat{i} + \hat{j} + 0\hat{k})\)
\(= 3\lambda + \mu + 0 = 10\)
Therefore:
\(3\lambda + \mu = 10\) ... (Equation 2)
We have two equations:
Substitute Equation 1 into Equation 2:
\(3(3\mu + 15) + \mu = 10\)
\(9\mu + 45 + \mu = 10\)
\(10\mu = 10 - 45\)
\(10\mu = -35\)
\(\mu = -3.5\)
Substitute \(\mu = -3.5\) into Equation 1:
\(\lambda = 3(-3.5) + 15\)
\(\lambda = -10.5 + 15\)
\(\lambda = 4.5\)
Finally, compute \(\lambda + \mu:\)
\(\lambda + \mu = 4.5 - 3.5 = 1\)
Thus, the value of \(\lambda + \mu\) is 1.