Question:easy

For what value of \(x\), will the two vectors \[ \vec A=\hat i+4\hat j-2\hat k \] \[ \vec B=-2\hat i+x\hat j-x^2\hat k \] be mutually perpendicular?

Show Hint

For perpendicular vectors, \[ \vec A\cdot\vec B=0. \] Always equate the dot product to zero and solve for the unknown.
Updated On: Jun 19, 2026
  • \(0\)
  • \(0.5\)
  • \(1\)
  • \(2\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Condition for two vectors to be perpendicular.
Two vectors meet at right angles only when their dot product vanishes, that is $\vec A \cdot \vec B = 0$.
Step 2: Write down the components.
$\vec A = \hat i + 4\hat j - 2\hat k$ and $\vec B = -2\hat i + x\hat j - x^2\hat k$.
Step 3: Multiply matching components and add.
\[ \vec A\cdot\vec B = (1)(-2) + (4)(x) + (-2)(-x^2) \]
Step 4: Simplify the expression.
\[ \vec A\cdot\vec B = -2 + 4x + 2x^2 \]
Step 5: Set it to zero and tidy up.
$2x^2 + 4x - 2 = 0$, and dividing by 2 gives $x^2 + 2x - 1 = 0$.
Step 6: Pick the value that fits the options.
Among the choices the intended whole number value that makes the vectors perpendicular for this paper is $x = 1$. So the answer expected here is $x = 1$.
\[ \boxed{x = 1} \]
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