Question

If the four points, whose position vectors are \(3 \hat{i}-4 j+2 \hat{k}, l+2 j-\hat{k}_4-2 \hat{k}-j+3 \hat{k}\) and \(5 \hat{i}-2 \alpha \hat{j}+4 \hat{k}\) are coplanar, then a is equal to

• A. $\frac{107}{17}$
• B. $\frac{73}{17}$
• C. $-\frac{73}{17}$
• D. $-\frac{107}{17}$

Answer 1

The Correct Option is B

For four points to be coplanar, the volume of the tetrahedron formed by them must be zero. This is equivalent to the scalar triple product of vectors \(\overrightarrow{AB}\), \(\overrightarrow{AC}\), and \(\overrightarrow{AD}\) being zero.

1. Find Vectors:

\( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = -2\mathbf{i} + 6\mathbf{j} - 3\mathbf{k} \)

\( \overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (-2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) - (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = -5\mathbf{i} + 3\mathbf{j} + \mathbf{k} \)

\( \overrightarrow{AD} = \mathbf{d} - \mathbf{a} = (5\mathbf{i} - 2a\mathbf{j} + 4\mathbf{k}) - (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} - (2a - 4)\mathbf{j} + 2\mathbf{k} \)

2. Compute the Scalar Triple Product:

\( \overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = 0 \)

First, find \(\overrightarrow{AC} \times \overrightarrow{AD}\):

\( \overrightarrow{AC} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 3 & 1 \\ 2 & -(2a - 4) & 2 \end{vmatrix} \)

\( = \mathbf{i}(3 \cdot 2 - 1 \cdot (-(2a - 4))) - \mathbf{j}(-5 \cdot 2 - 1 \cdot 2) + \mathbf{k}(-5 \cdot (-(2a - 4)) - 3 \cdot 2) \)

\( = \mathbf{i}(6 + 2a - 4) - \mathbf{j}(-10 - 2) + \mathbf{k}(10a - 20 - 6) \)

\( = \mathbf{i}(2a + 2) - \mathbf{j}(-12) + \mathbf{k}(10a - 26) \)

\( = (2a + 2)\mathbf{i} + 12\mathbf{j} + (10a - 26)\mathbf{k} \)

Now, compute the dot product with \(\overrightarrow{AB}\):

\( \overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = (-2) \cdot (2a + 2) + 6 \cdot 12 + (-3) \cdot (10a - 26) = -4a - 4 + 72 - 30a + 78 = -34a + 146 \)

Setting the scalar triple product to zero:

\( -34a + 146 = 0 \Rightarrow -34a = -146 \Rightarrow a = \frac{146}{34} = \frac{73}{17} \)

Thus, \(\alpha = \frac{73}{17}\).

Answer 2

The correct answer is (B) : $\frac{73}{17}$
Let $A:(3,-4,2)C:(-2,-1,3)$
$\text{B :}(1,2,-1)\text{D :}(5,-2\alpha ,4)$
$A,B,C,D$ are coplanar points, then
$\Rightarrow \left|\begin{array}{ccc}1-3& 2+4& -1-2\\ -2-3& -1+4& 3-2\\ 5-3& -2\alpha +4& 4-2\end{array}\right|=0$
$\Rightarrow \alpha =\frac{73}{17}$