Understanding Coplanar Vectors

Three vectors are considered coplanar vectors if they lie on the same plane. In three-dimensional space (\(3D vectors\)), a fundamental condition for three vectors to be coplanar is that their scalar triple product must be zero. The scalar triple product of three vectors, say \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), is given by \(\vec{a} \cdot (\vec{b} \times \vec{c})\).

Alternatively, the scalar triple product can be calculated as the determinant of the matrix formed by the components of the three vectors. For vectors \(\vec{a} = a_1 \widehat i + a_2 \widehat j + a_3 \widehat k\), \(\vec{b} = b_1 \widehat i + b_2 \widehat j + b_3 \widehat k\), and \(\vec{c} = c_1 \widehat i + c_2 \widehat j + c_3 \widehat k\), their scalar triple product is:

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

For the vectors to be coplanar vectors, this determinant must be equal to zero. This concept is a key part of vector algebra.

Applying the Coplanarity Condition

We are given three vectors:

• \(\vec{v}_1 = \lambda \widehat i + \widehat j + 2\widehat k\)
• \(\vec{v}_2 = \widehat i + \lambda \widehat j - \widehat k\)
• \(\vec{v}_3 = 2\widehat i - \widehat j + \lambda \widehat k\)

For these vectors to be coplanar vectors, their scalar triple product must be zero. We can set up the determinant using their components:

\[ \begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{vmatrix} = 0 \]

Now, we need to evaluate this determinant and solve for \(\lambda\). This is a typical step in finding lambda for such problems.

Evaluating the Determinant

We can expand the determinant along the first row:

\[ \lambda \begin{vmatrix} \lambda & -1 \\ -1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 2 & \lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} = 0 \]

Evaluate the \(2 \times 2\) determinants:

• \(\begin{vmatrix} \lambda & -1 \\ -1 & \lambda \end{vmatrix} = (\lambda)(\lambda) - (-1)(-1) = \lambda^2 - 1\)
• \(\begin{vmatrix} 1 & -1 \\ 2 & \lambda \end{vmatrix} = (1)(\lambda) - (-1)(2) = \lambda + 2\)
• \(\begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} = (1)(-1) - (\lambda)(2) = -1 - 2\lambda\)

Substitute these back into the expanded equation:

\[ \lambda (\lambda^2 - 1) - 1 (\lambda + 2) + 2 (-1 - 2\lambda) = 0 \]

Solving for \(\lambda\)

Now, simplify and solve the equation for \(\lambda\). This is the final step in finding lambda.

\[ \lambda^3 - \lambda - \lambda - 2 - 2 - 4\lambda = 0 \]

Combine like terms:

\[ \lambda^3 - 6\lambda - 4 = 0 \]

We need to find the roots of this cubic equation. We can test simple integer values like -2, -1, 0, 1, 2.

• If \(\lambda = -2\): \((-2)^3 - 6(-2) - 4 = -8 + 12 - 4 = 0\). So, \(\lambda = -2\) is a root.

Since \(\lambda = -2\) is a root, \(( \lambda + 2 )\) is a factor of the polynomial \(\lambda^3 - 6\lambda - 4\). We can perform polynomial division or synthetic division to find other factors.

Using synthetic division with root -2:

The resulting quadratic is \(\lambda^2 - 2\lambda - 2 = 0\). We can find the roots of this quadratic using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ \lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \]

So, the roots of the cubic equation are \(\lambda = -2\), \(\lambda = 1 + \sqrt{3}\), and \(\lambda = 1 - \sqrt{3}\). These are the values of \(\lambda\) for which the vectors are coplanar vectors.

Conclusion

The values of \(\lambda\) that make the given three vectors coplanar are \(-2\), \(1 + \sqrt{3}\), and \(1 - \sqrt{3}\). One of the options provided is \(-2\).

Therefore, the vectors are coplanar if \(\lambda = -2\).