Question

If \(\lambda(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\) is a unit vector, then the value(s) of \(\lambda\) are:

• A. \(\pm \dfrac{1}{7}\)
• B. \(\pm 7\)
• C. \(\pm \sqrt{43}\)
• D. \(\pm \dfrac{1}{\sqrt{43}}\)

Hint:

To convert any vector (vecv) into a unit vector, divide it by its magnitude: [ hatv = fracvecv|vecv| ]

Answer 1

The Correct Option is D

Step 1: Recall the definition of a unit vector. A vector is a unit vector if its magnitude is equal to \(1\). Step 2: Find the magnitude of the given vector. Given vector: \[ \lambda(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \] Magnitude: \[ \left|\lambda(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\right| = |\lambda|\sqrt{3^2 + 2^2 + (-2)^2} \] \[ = |\lambda|\sqrt{9 + 4 + 4} = |\lambda|\sqrt{17} \] Step 3: Apply the unit vector condition. \[ |\lambda|\sqrt{17} = 1 \] \[ |\lambda| = \frac{1}{\sqrt{17}} \] Step 4: Write both possible values of \(\lambda\). \[ \lambda = \pm \frac{1}{\sqrt{17}} \] Step 5: Final conclusion. The correct answer is: \[ \boxed{\pm \dfrac{1}{\sqrt{17}}} \]