Understanding Coplanarity of Vectors

Three vectors are considered coplanar if they lie in the same plane. A common method to determine if three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ are coplanar is to check their scalar triple product. If the scalar triple product is zero, the vectors are coplanar.

The scalar triple product of three vectors $\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$, $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$, and $\vec{w} = w_1\hat{i} + w_2\hat{j} + w_3\hat{k}$ is given by the determinant of the matrix formed by their components:

$ \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:

$ \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} = 0 $

Applying Coplanarity Condition to Given Vectors

The given vectors are:

• $\vec{u} = 2\hat{i} - \hat{j} + \hat{k}$ (Components: $u_1=2, u_2=-1, u_3=1$)
• $\vec{v} = \hat{i} + 2\hat{j} - 3\hat{k}$ (Components: $v_1=1, v_2=2, v_3=-3$)
• $\vec{w} = 3\hat{i} + a\hat{j} + 5\hat{k}$ (Components: $w_1=3, w_2=a, w_3=5$)

To find the value of 'a' for which these vectors are coplanar, we set up the determinant of their components and set it equal to zero:

$ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & a & 5 \end{vmatrix} = 0 $

Solving the Determinant for 'a'

We can evaluate the determinant by expanding along the first row:

$ 2 \begin{vmatrix} 2 & -3 \\ a & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & a \end{vmatrix} = 0 $

Now, we evaluate the $2 \times 2$ determinants:

• $ \begin{vmatrix} 2 & -3 \\ a & 5 \end{vmatrix} = (2)(5) - (-3)(a) = 10 - (-3a) = 10 + 3a $
• $ \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} = (1)(5) - (-3)(3) = 5 - (-9) = 5 + 9 = 14 $
• $ \begin{vmatrix} 1 & 2 \\ 3 & a \end{vmatrix} = (1)(a) - (2)(3) = a - 6 $

Substitute these values back into the expansion equation:

$ 2(10 + 3a) + 1(14) + 1(a - 6) = 0 $

Now, simplify and solve for 'a':

$ 20 + 6a + 14 + a - 6 = 0 $

Combine the terms with 'a' and the constant terms:

$ (6a + a) + (20 + 14 - 6) = 0 $

$ 7a + (34 - 6) = 0 $

$ 7a + 28 = 0 $

Subtract 28 from both sides:

$ 7a = -28 $

Divide by 7:

$ a = \frac{-28}{7} $

$ a = -4 $

Thus, the value of 'a' that makes the three vectors coplanar is -4.

Revision Table: Vector Coplanarity Concepts

Additional Information on Vector Coplanarity and Scalar Triple Product

The scalar triple product $ \vec{u} \cdot (\vec{v} \times \vec{w}) $ can be interpreted geometrically as the volume of the parallelepiped formed by the three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ as adjacent edges. If the three vectors are coplanar, they cannot form a parallelepiped with a non-zero volume. Hence, the volume must be zero, which corresponds to the scalar triple product being zero.

The order of the vectors in the scalar triple product matters for the sign of the result, but not for whether the product is zero. For example, $ \vec{u} \cdot (\vec{v} \times \vec{w}) = - \vec{v} \cdot (\vec{u} \times \vec{w}) $. However, for coplanarity, we only care if the result is zero, so the order within the determinant can be permuted cyclically without changing the zero condition.

Another way to think about coplanarity is that one of the vectors can be expressed as a linear combination of the other two. That is, if $\vec{u}$, $\vec{v}$, and $\vec{w}$ are coplanar and $\vec{v}$ and $\vec{w}$ are not parallel, then $\vec{u} = c_1 \vec{v} + c_2 \vec{w}$ for some scalars $c_1$ and $c_2$. This linear dependence is equivalent to the scalar triple product being zero.