Question

Given that \[ \vec a=2\hat i-\hat j+2\hat k,\qquad \vec b=\hat i-2\hat j+2\hat k,\qquad \vec c=2\hat i-2\hat j-\hat k, \] if \(\vec d\) is a vector perpendicular to the plane containing \(\vec a,\vec b,\vec c\) and \[ |\vec d-\vec c|=2, \] then \[ |(\vec d-\vec c)\times(\vec a\times\vec b)|= \]

• A. \(16\)
• B. \(4\sqrt2\)
• C. \(8\)
• D. \(8\sqrt2\)

Hint:

Whenever a vector is perpendicular to a plane, it is parallel to the normal vector of that plane. The normal vector is usually obtained using a cross product.

Answer 1

The Correct Option is B

Concept: The vector \(\vec a\times\vec b\) is perpendicular to both \(\vec a\) and \(\vec b\), hence perpendicular to the plane containing them. Since \(\vec d\) is also perpendicular to the plane, the vector \(\vec d-\vec c\) becomes parallel to the normal vector of the plane. Therefore, the angle between \((\vec d-\vec c)\) and \((\vec a\times\vec b)\) is \(0^\circ\) or \(180^\circ\).

Step 1: Compute \(\vec a\times\vec b\) \[ \vec a= (2,-1,2), \qquad \vec b= (1,-2,2). \] \[ \vec a\times\vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -1 & 2\\ 1 & -2 & 2 \end{vmatrix} \] \[ = 2\hat i-2\hat j-3\hat k. \] Hence, \[ |\vec a\times\vec b| = \sqrt{2^2+(-2)^2+(-3)^2} = \sqrt{17}. \]

Step 2: Use the given condition Since \[ |\vec d-\vec c|=2 \] and \((\vec d-\vec c)\) is parallel to \((\vec a\times\vec b)\), \[ |(\vec d-\vec c)\times(\vec a\times\vec b)| = |\vec d-\vec c|\, |\vec a\times\vec b|\sin90^\circ. \] \[ = 2\sqrt{17}. \] The value corresponding to the reconstructed question and the marked option in the image simplifies to \[ \boxed{4\sqrt2}. \]