Let a = i + j + k, b = i - j + 2k and c = xi + (x - 2)j - k. If the vectors c lies in the plane of a

Q: Why does the scalar triple product equal zero? Isn't it just a number?

It is indeed a scalar quantity, but that scalar represents the volume of the parallelepiped formed by the three vectors. If the volume is zero, it implies that the vectors are linearly dependent and cannot span a 3D space; they are effectively trapped on a 2D plane.

Q: Can I use the cross product method instead of the determinant?

Absolutely! You can calculate $\vec{n} = \vec{a} \times \vec{b}$, which gives you a vector normal to the plane containing $\vec{a}$ and $\vec{b}$. If $\vec{c}$ is coplanar, it must be perpendicular to this normal vector. Therefore, the condition is $\vec{n} \cdot \vec{c} = 0$. It is mathematically equivalent to the determinant method.

Q: I keep messing up the signs in the determinant expansion. Any tips?

The cofactor expansion rule is $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei-fh) - b(di-fg) + c(dh-eg)$. Always write the signs $+ - +$ above the first row before you start to avoid the classic sign-flip error!

Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$. If the vectors $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then $x$ equals

Options:

(A) $-4$

(B) $-2$

(C) 0

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To determine the value of $x$ such that the vector $\vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$ lies in the plane defined by $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$, we must invoke the condition of **coplanarity**.

Geometrically, three vectors are coplanar if the volume of the parallelepiped they form is zero. This implies that the vectors are linearly dependent and do not span a three-dimensional volume.

### The Scalar Triple Product

The condition for coplanarity is expressed mathematically using the **Scalar Triple Product**, denoted as $[\vec{a} \ \vec{b} \ \vec{c}]$. We set this product equal to zero:

$$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = 0$$

This operation is most efficiently computed using the determinant of a $3 	imes 3$ matrix, where each row consists of the components of the respective vectors.

### The Determinant Formulation

We arrange the components of $\vec{a}$, $\vec{b}$, and $\vec{c}$ into the following determinant:

$$\begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & 2 \ x & x-2 & -1 \end{vmatrix} = 0$$

To solve for $x$, we expand this determinant along the first row using the standard cofactor expansion method.

### The Expansion Process

Expanding along the first row, we calculate the minors for each element:

$$1 \cdot \begin{vmatrix} -1 & 2 \ x-2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 2 \ x & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \ x & x-2 \end{vmatrix} = 0$$

Evaluating these $2 	imes 2$ determinants individually:

$$1 \cdot [(-1)(-1) - (2)(x-2)] = 1 - 2x + 4 = 5 - 2x$$

$$-1 \cdot [(1)(-1) - (2)(x)] = -1 \cdot [-1 - 2x] = 1 + 2x$$

$$1 \cdot [(1)(x-2) - (-1)(x)] = x - 2 + x = 2x - 2$$

### Final Calculation

Summing these results to satisfy the determinant equation:

$$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$$

Simplifying the expression by combining like terms:

$$2x + 4 = 0$$

Solving for $x$, we find:

$$2x = -4 \Rightarrow **x = -2**$$

When **$x = -2$**, the vector $\vec{c}$ is perfectly coplanar with vectors $\vec{a}$ and $\vec{b}$.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = x \hat{i} + (x - 2) \hat{j} - \hat{k}$ . If the vectors $c$ lies in the plane of $a$ and $b$ , then $x$ equals

Select Answer:

Visualized Solution

Visualizing the Vectors

Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant: Part 1

Expanding the Determinant: Part 2

Expanding the Determinant: Part 3

Combining the Terms

Simplifying the Equation

Solving for $x$

Final Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Scalar Triple Product

The Determinant Formulation

The Expansion Process

Final Calculation

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