Question

If $\vec{a} = 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}$, $\vec{b} = \hat{\text{i}} + 2\hat{\text{j}} - 3\hat{\text{k}}$ and $\vec{c} = 3\hat{\text{i}} + \lambda\hat{\text{j}} + 5\hat{\text{k}}$ are coplanar, then $\lambda$ is the root of the equation

• A. $x^2 + 3x = 6$
• B. $x^2 + 2x = 4$
• C. $x^2 + 3x = 4$
• D. $x^2 + 2x = 6$

Hint:

Once you solve for your linear variable value ($\lambda = -4$), quickly plugging it into the options is much faster than attempting to algebraically factor or expand each option choice from scratch!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given three vectors containing an unknown component $\lambda$. We are told that these vectors are coplanar, and we need to determine which quadratic equation has $\lambda$ as a valid mathematical root.

Step 2: Key Formula or Approach:
Three vectors are coplanar if and only if their scalar triple product is equal to zero. This is calculated by setting the determinant of their component coefficients to zero:
$$\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$$

Step 3: Detailed Explanation:
Set up the matrix determinant using the given vector components:
$$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} = 0$$ Expand the determinant along the first row:
$$2\begin{vmatrix} 2 & -3 \\ \lambda & 5 \end{vmatrix} - (-1)\begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} + 1\begin{vmatrix} 1 & 2 \\ 3 & \lambda \end{vmatrix} = 0$$ Evaluate the $2 \times 2$ minor blocks:
$$2(10 - (-3\lambda)) + 1(5 - (-9)) + 1(\lambda - 6) = 0$$ $$2(10 + 3\lambda) + 1(14) + (\lambda - 6) = 0$$ Expand and combine like terms:
$$20 + 6\lambda + 14 + \lambda - 6 = 0$$ $$7\lambda + 28 = 0 \implies 7\lambda = -28 \implies \lambda = -4$$ Now, let's test which of the given quadratic equations holds true when we plug in $x = \lambda = -4$:
For option (A): $(-4)^2 + 3(-4) = 16 - 12 = 4 \neq 6$ (False)
For option (B): $(-4)^2 + 2(-4) = 16 - 8 = 8 \neq 4$ (False)
For option (C): $(-4)^2 + 3(-4) = 16 - 12 = 4 = 4$ (True)
Therefore, $\lambda = -4$ is a root of the equation $x^2 + 3x = 4$.

Step 4: Final Answer:
The value of $\lambda$ satisfies the equation $x^2 + 3x = 4$, which corresponds to option (C).