Understanding Coplanarity of Vectors

Three vectors are said to be coplanar if they lie in the same plane. A fundamental condition for three vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) to be coplanar is that their scalar triple product is zero. The scalar triple product can be calculated as the determinant of the matrix formed by their components.

The scalar triple product of \(\vec{u} = u_1\hat i + u_2\hat j + u_3\hat k\), \(\vec{v} = v_1\hat i + v_2\hat j + v_3\hat k\), and \(\vec{w} = w_1\hat i + w_2\hat j + w_3\hat k\) is given by:

\([\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}\)

For the vectors to be coplanar, this determinant must be equal to zero.

Setting up the Determinant for Coplanar Vectors

The given vectors are:

• \(\vec{v}_1 = a\hat i + \hat j + \hat k\), with components \(\langle a, 1, 1 \rangle\)
• \(\vec{v}_2 = \hat i + b\hat j + \hat k\), with components \(\langle 1, b, 1 \rangle\)
• \(\vec{v}_3 = \hat i + \hat j + c\hat k\), with components \(\langle 1, 1, c \rangle\)

Since these vectors are coplanar, their scalar triple product is zero. We can write this as the determinant of the matrix formed by their components:

\( \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \)

Evaluating the Determinant and Deriving the Coplanarity Equation

Now, we evaluate the determinant. Expanding along the first row, we get:

\( a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} = 0 \)

\( a(bc - 1) - 1(c - 1) + 1(1 - b) = 0 \)

Simplify the expression:

\( abc - a - (c - 1) + (1 - b) = 0 \)

\( abc - a - c + 1 + 1 - b = 0 \)

\( abc - a - b - c + 2 = 0 \)

This is the condition that must be satisfied for the given vectors to be coplanar. We can rearrange this equation as:

\( abc - (a + b + c) + 2 = 0 \)

Evaluating the Expression \(\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}\)

We need to find the value of the expression \(\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}\). Let's combine these terms by finding a common denominator, which is \((1 - a)(1 - b)(1 - c)\). Note that \(a, b, c \ne 1\) ensures the denominators are non-zero.

\( \frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} = \frac{(1 - b)(1 - c)}{(1 - a)(1 - b)(1 - c)} + \frac{(1 - a)(1 - c)}{(1 - a)(1 - b)(1 - c)} + \frac{(1 - a)(1 - b)}{(1 - a)(1 - b)(1 - c)} \)

Combine the numerators over the common denominator:

\( = \frac{(1 - b)(1 - c) + (1 - a)(1 - c) + (1 - a)(1 - b)}{(1 - a)(1 - b)(1 - c)} \)

Expand the terms in the numerator:

• \((1 - b)(1 - c) = 1 - c - b + bc\)
• \((1 - a)(1 - c) = 1 - c - a + ac\)
• \((1 - a)(1 - b) = 1 - b - a + ab\)

Summing the numerators:

\( (1 - c - b + bc) + (1 - c - a + ac) + (1 - b - a + ab) \)

\( = 1 - c - b + bc + 1 - c - a + ac + 1 - b - a + ab \)

\( = (1 + 1 + 1) + (-a - a) + (-b - b) + (-c - c) + ab + bc + ac \)

\( = 3 - 2a - 2b - 2c + ab + bc + ac \)

\( = 3 - 2(a + b + c) + (ab + bc + ac) \)

Now expand the denominator:

\( (1 - a)(1 - b)(1 - c) = (1 - a)(1 - c - b + bc) \)

\( = 1(1 - c - b + bc) - a(1 - c - b + bc) \)

\( = 1 - c - b + bc - a + ac + ab - abc \)

\( = 1 - (a + b + c) + (ab + bc + ac) - abc \)

So the expression becomes:

\( \frac{3 - 2(a + b + c) + (ab + bc + ac)}{1 - (a + b + c) + (ab + bc + ac) - abc} \)

From the coplanarity condition, we have \(abc - (a + b + c) + 2 = 0\). We can rearrange this as \(abc = a + b + c - 2\). Let's see if substituting this simplifies the fraction. This approach seems complicated. Let's try substituting the target value (1) into the expression and see if it matches the coplanarity condition.

Assume \(\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} = 1\). Then:

\( \frac{(1 - b)(1 - c) + (1 - a)(1 - c) + (1 - a)(1 - b)}{(1 - a)(1 - b)(1 - c)} = 1 \)

\( (1 - b)(1 - c) + (1 - a)(1 - c) + (1 - a)(1 - b) = (1 - a)(1 - b)(1 - c) \)

Substitute the expanded forms we found earlier:

\( [1 - c - b + bc] + [1 - c - a + ac] + [1 - b - a + ab] = [1 - (a + b + c) + (ab + bc + ac) - abc] \)

\( 3 - 2(a + b + c) + (ab + bc + ac) = 1 - (a + b + c) + (ab + bc + ac) - abc \)

Now, move all terms to one side to see if we get the coplanarity equation:

\( 3 - 2(a + b + c) + (ab + bc + ac) - [1 - (a + b + c) + (ab + bc + ac) - abc] = 0 \)

\( 3 - 2(a + b + c) + (ab + bc + ac) - 1 + (a + b + c) - (ab + bc + ac) + abc = 0 \)

Combine like terms:

\( (3 - 1) + (-2(a + b + c) + (a + b + c)) + ((ab + bc + ac) - (ab + bc + ac)) + abc = 0 \)

\( 2 - (a + b + c) + 0 + abc = 0 \)

\( abc - (a + b + c) + 2 = 0 \)

This is exactly the equation we derived from the coplanarity condition. Therefore, the assumption that the expression equals 1 is correct, as it leads directly to the condition for the vectors being coplanar.

Final Answer Calculation

Based on the derivation from the coplanarity condition, the value of the expression \(\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}\) is 1.

Additional Information on Coplanar Vector Properties

Understanding coplanar vectors is crucial in 3D geometry and vector algebra. Here are some additional points:

• Linear Dependence: Three vectors \(\vec{u}, \vec{v}, \vec{w}\) are coplanar if and only if one of them can be expressed as a linear combination of the other two. That is, \(\vec{w} = p\vec{u} + q\vec{v}\) for some scalars \(p\) and \(q\).
• Geometric Interpretation: If three vectors are drawn from the same origin, they are coplanar if their endpoints lie on a single plane passing through the origin.
• Non-Collinear Vectors: Any two non-collinear vectors are always coplanar; they define a plane. Coplanarity becomes a relevant condition when considering three or more vectors.
• Applications: Coplanarity is used in physics to determine if forces or velocities lie in the same plane, simplifying analysis. It's also used in geometry to check if points lie on the same plane or if lines intersect.