Key Concepts and Formulas

• Collinear Vectors: Two non-zero vectors $\overrightarrow A$ and $\overrightarrow B$ are collinear if and only if $\overrightarrow A = k\overrightarrow B$ for some non-zero scalar $k$. This implies that their corresponding components are proportional.
• Equality of Vectors: Two vectors are equal if and only if their corresponding components are equal.
• Unit Vector: A unit vector in the direction of a non-zero vector $\overrightarrow V$ is given by $\widehat V = \frac{\overrightarrow V}{|\overrightarrow V|}$, where $|\overrightarrow V|$ is the magnitude of $\overrightarrow V$. The magnitude of a vector $\overrightarrow V = a\widehat i + b\widehat j + c\widehat k$ is $|\overrightarrow V| = \sqrt{a^2 + b^2 + c^2}$.

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Step-by-Step Solution

Step 1: Expressing Collinearity Mathematically

We are given two vectors, $\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$ and $\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$. Since $\overrightarrow {{a_1}}$ and $\overrightarrow {{a_2}}$ are collinear, there exists a non-zero scalar $\lambda$ such that $\overrightarrow {{a_2}} = \lambda \overrightarrow {{a_1}}$.

$\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)$ $\widehat i + y\widehat j + z\widehat k = (\lambda x)\widehat i - \lambda\widehat j + \lambda\widehat k$

Why this step? This is the definition of collinear vectors, which allows us to establish a relationship between the two given vectors and introduce a scalar parameter $\lambda$.

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Step 2: Equating Components to Find Relationships

For the equality of vectors to hold, their corresponding components must be equal. Equating the coefficients of $\widehat i$, $\widehat j$, and $\widehat k$ on both sides of the equation from Step 1:

1. Coefficient of $\widehat i$: $1 = \lambda x$
2. Coefficient of $\widehat j$: $y = -\lambda$
3. Coefficient of $\widehat k$: $z = \lambda$

From these equations, we can express $x, y, z$ in terms of $\lambda$:

• From $1 = \lambda x$, we get $x = \frac{1}{\lambda}$. (Note: Since $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ are non-zero and collinear, $\lambda$ cannot be zero.)
• $y = -\lambda$.
• $z = \lambda$.

Why this step? By equating components, we derive the specific relationships between the unknown variables ($x, y, z$) and the scalar parameter ($\lambda$). This is crucial for substituting these relationships into the vector we are interested in.

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Step 3: Constructing the Target Vector

We need to find a unit vector parallel to the vector $x\widehat i + y\widehat j + z\widehat k$. Let this vector be $\overrightarrow V$. $\overrightarrow V = x\widehat i + y\widehat j + z\widehat k$ Substitute the expressions for $x, y, z$ found in Step 2 into $\overrightarrow V$: $\overrightarrow V = \left(\frac{1}{\lambda}\right)\widehat i + (-\lambda)\widehat j + (\lambda)\widehat k$ $\overrightarrow V = \frac{1}{\lambda}\widehat i - \lambda\widehat j + \lambda\widehat k$

Why this step? We have now expressed the target vector $\overrightarrow V$ in terms of a single scalar parameter $\lambda$. This simplifies the problem, as we can now analyze the direction of $\overrightarrow V$ based on the value of $\lambda$.

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Step 4: Finding a Unit Vector Parallel to $\overrightarrow V$

A unit vector in the direction of $\overrightarrow V$ is given by $\frac{\overrightarrow V}{|\overrightarrow V|}$. First, let's find the magnitude of $\overrightarrow V$: $|\overrightarrow V| = \left| \frac{1}{\lambda}\widehat i - \lambda\widehat j + \lambda\widehat k \right| = \sqrt{\left(\frac{1}{\lambda}\right)^2 + (-\lambda)^2 + (\lambda)^2}$ $|\overrightarrow V| = \sqrt{\frac{1}{\lambda^2} + \lambda^2 + \lambda^2} = \sqrt{\frac{1}{\lambda^2} + 2\lambda^2}$

The unit vector $\widehat u$ is: $\widehat u = \frac{\overrightarrow V}{|\overrightarrow V|} = \frac{\frac{1}{\lambda}\widehat i - \lambda\widehat j + \lambda\widehat k}{\sqrt{\frac{1}{\lambda^2} + 2\lambda^2}}$ To simplify, multiply the numerator and the denominator by $\lambda$ (since $\lambda \neq 0$): $\widehat u = \frac{\lambda \left(\frac{1}{\lambda}\widehat i - \lambda\widehat j + \lambda\widehat k\right)}{\lambda \sqrt{\frac{1}{\lambda^2} + 2\lambda^2}} = \frac{\widehat i - \lambda^2\widehat j + \lambda^2\widehat k}{\sqrt{\lambda^2 \left(\frac{1}{\lambda^2} + 2\lambda^2\right)}}$ $\widehat u = \frac{\widehat i - \lambda^2\widehat j + \lambda^2\widehat k}{\sqrt{1 + 2\lambda^4}}$

Why this step? This step applies the definition of a unit vector. By calculating the magnitude and dividing the vector by it, we obtain a vector of unit length pointing in the same direction as $\overrightarrow V$. The simplification makes it easier to compare with the given options.

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Step 5: Choosing a Convenient Value for $\lambda$

The problem asks for "a possible unit vector". This means we can choose any non-zero value for $\lambda$ that simplifies the expression for $\widehat u$ and matches one of the options. A common and effective choice is to set $\lambda^2 = 1$, which means $\lambda = 1$ or $\lambda = -1$. Let's choose $\lambda^2 = 1$.

Substitute $\lambda^2 = 1$ into the expression for $\widehat u$: $\widehat u = \frac{\widehat i - (1)\widehat j + (1)\widehat k}{\sqrt{1 + 2(1)^2}}$ $\widehat u = \frac{\widehat i - \widehat j + \widehat k}{\sqrt{1 + 2}}$ $\widehat u = \frac{\widehat i - \widehat j + \widehat k}{\sqrt{3}}$ This can be written as: $\widehat u = \frac{1}{\sqrt{3}}(\widehat i - \widehat j + \widehat k)$

Why this step? Since we need to find a possible unit vector, we can pick a value for $\lambda$ that leads to a simple form. Setting $\lambda^2=1$ is strategic because it simplifies terms involving $\lambda^2$ and $\lambda^4$, making the magnitude calculation cleaner. This resulting vector matches option (A).

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Common Mistakes & Tips

• Proportionality of Components: When vectors are collinear, not only are the vectors proportional, but so are their corresponding components. For example, if $\overrightarrow A = k\overrightarrow B$, then $A_x = kB_x$, $A_y = kB_y$, and $A_z = kB_z$.
• Scalar $\lambda$ can be negative: Remember that the scalar multiple $\lambda$ can be positive or negative, indicating the direction of collinearity.
• Magnitude Calculation: Be careful when squaring terms and taking square roots, especially when variables are involved.
• Unit Vector Definition: Always divide the vector by its magnitude to obtain a unit vector.

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Summary

The problem hinges on the definition of collinear vectors, which states that one vector can be expressed as a scalar multiple of another. By equating the components of the given collinear vectors, we established relationships between their unknown variables ($x, y, z$) and a scalar parameter ($\lambda$). These relationships allowed us to express the target vector $x\widehat i + y\widehat j + z\widehat k$ in terms of $\lambda$. We then calculated the general form of a unit vector parallel to this target vector. By strategically choosing a simple value for $\lambda$ (specifically, $\lambda^2=1$), we obtained a unit vector that matches one of the provided options.

The final answer is $\boxed{\text{(A)}}$.