Coplanarity of three vectors

Coplanarity of Three Vectors

Three vectors \( \vec{a}, \vec{b}, \vec{c} \) are said to be coplanar if they lie in the same plane. This means that one of the vectors can be expressed as a linear combination of the other two, i.e.,

\[ \vec{a}, \vec{b}, \vec{c} \; are\; coplanar\; if } \vec{a} = x\vec{b} + y\vec{c} \; for\; some\; scalars }\; x,\; y. \]

Another approach is to check whether the scalar triple product of the three vectors equals zero.

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0. \]

1) Show that \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}, \vec{c} = 3\hat{i} - 4\hat{j} - 4\hat{k} \) are coplanar.

Solve:

We will show that \( \vec{a} = x\vec{b} + y\vec{c} \) for some scalars \( x \) and \( y \).

Let us assume:

\[ \vec{a} = x\vec{b} + y\vec{c} \]

Substituting the values:

\[ 2\hat{i} - \hat{j} + \hat{k} = x(\hat{i} - 3\hat{j} - 5\hat{k}) + y(3\hat{i} - 4\hat{j} - 4\hat{k}) \]

Expanding the right-hand side:

\[ = x\hat{i} - 3x\hat{j} - 5x\hat{k} + 3y\hat{i} - 4y\hat{j} - 4y\hat{k} \]

= \( (x + 3y)\hat{i} + (-3x - 4y)\hat{j} + (-5x - 4y)\hat{k} \)

Now compare coefficients:

• \( x + 3y = 2 \) — (i)
• \( -3x - 4y = -1 \) — (ii)
• \( -5x - 4y = 1 \) — (iii)

From (i):

\[ x = 2 - 3y \]

Substitute in (ii):

\[ -3(2 - 3y) - 4y = -1 \\ -6 + 9y - 4y = -1 \\ 5y = 5 \\ y = 1 \]

Then, \( x = 2 - 3(1) = -1 \)

Now check in (iii):

\[ -5(-1) - 4(1) = 5 - 4 = 1 \]

So all three equations are satisfied.

Hence, the given vectors are coplanar.

2) Prove that four points A, B, C, D whose position vectors are \( 2\hat{i} + 3\hat{j} - \hat{k}, \hat{i} - 2\hat{j} + 3\hat{k}, 23\hat{i} + 4\hat{j} - 2\hat{k}, \hat{i} - 6\hat{j} + \hat{k} \) are coplanar.

Solve:

Let the position vectors be:

• \( \vec{A} = 2\hat{i} + 3\hat{j} - \hat{k} \)
• \( \vec{B} = \hat{i} - 2\hat{j} + 3\hat{k} \)
• \( \vec{C} = 23\hat{i} + 4\hat{j} - 2\hat{k} \)
• \( \vec{D} = \hat{i} - 6\hat{j} + \hat{k} \)

We need to show vectors \( \vec{AB}, \vec{AC}, \vec{AD} \) are coplanar, i.e., \( \vec{AD} = x\vec{AB} + y\vec{AC} \)

\[ \vec{AB} = \vec{B} - \vec{A} = (\hat{i} - 2\hat{j} + 3\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) = -\hat{i} - 5\hat{j} + 4\hat{k} \]

\[ \vec{AC} = \vec{C} - \vec{A} = (23\hat{i} + 4\hat{j} - 2\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) = 21\hat{i} + \hat{j} - \hat{k} \]

\[ \vec{AD} = \vec{D} - \vec{A} = (\hat{i} - 6\hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) = -\hat{i} - 9\hat{j} + 2\hat{k} \]

Assume \( \vec{AD} = x\vec{AB} + y\vec{AC} \)

\[ -\hat{i} - 9\hat{j} + 2\hat{k} = x(-\hat{i} - 5\hat{j} + 4\hat{k}) + y(21\hat{i} + \hat{j} - \hat{k}) \]

Expand and compare:

• \( -x + 21y = -1 \) … (1)
• \( -5x + y = -9 \) … (2)
• \( 4x - y = 2 \) … (3)

Solving (2) and (3):

From (3): \( y = 4x - 2 \)

Substitute in (2):

\[ -5x + (4x - 2) = -9 \\ -x - 2 = -9 \Rightarrow x = 7 \Rightarrow y = 4(7) - 2 = 26 \]

Check in (1):

-x + 21y = -7 + 546 = 539 ≠ -1

The contradiction indicates that there may be a miscalculation or a false assumption. Instead, use scalar triple product method:

\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0 \Rightarrow \;coplanar \]

But since vector approach failed to satisfy a consistent solution with all three equations, and scalar triple product is more suitable here, matrix method is preferable for verification.