Question

The vector \(\vec{x}\) is perpendicular to the vectors \(\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\), \(\vec{b}=18\hat{i}-22\hat{j}-5\hat{k}\) and makes an obtuse angle with \(\hat{j}\). If \(|\vec{x}|=14\), then \(\vec{x}=\)

• A. \(8\hat{i}+12\hat{j}+24\hat{k}\)
• B. \(-8\hat{i}+6\hat{j}+24\hat{k}\)
• C. \(8\hat{i}-12\hat{j}-24\hat{k}\)
• D. \(-8\hat{i}-12\hat{j}+24\hat{k}\)

Hint:

If a vector is perpendicular to two given vectors, then it is parallel to their cross product. Also, a vector makes an obtuse angle with \(\hat{j}\) when its \(j\)-component is negative.

Answer 1

The Correct Option is D

Step 1: Use the condition of perpendicularity.
Since \(\vec{x}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), \(\vec{x}\) must be parallel to \[ \vec{a}\times \vec{b} \] Now, \[ \vec{a}=3\hat{i}+2\hat{j}+2\hat{k} \] and \[ \vec{b}=18\hat{i}-22\hat{j}-5\hat{k} \]

Step 2: Find \(\vec{a}\times \vec{b}\).
\[ \vec{a}\times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 18 & -22 & -5 \end{vmatrix} \] \[ = \hat{i}\{2(-5)-2(-22)\} -\hat{j}\{3(-5)-2(18)\} +\hat{k}\{3(-22)-2(18)\} \] \[ = \hat{i}(-10+44) -\hat{j}(-15-36) +\hat{k}(-66-36) \] \[ = 34\hat{i}+51\hat{j}-102\hat{k} \] Thus, \[ \vec{a}\times \vec{b}=17(2\hat{i}+3\hat{j}-6\hat{k}) \] So, \(\vec{x}\) is parallel to \[ 2\hat{i}+3\hat{j}-6\hat{k} \]

Step 3: Use the magnitude condition.
Let \[ \vec{x}=\lambda(2\hat{i}+3\hat{j}-6\hat{k}) \] Then, \[ |\vec{x}|=|\lambda|\sqrt{2^2+3^2+(-6)^2} \] \[ =|\lambda|\sqrt{4+9+36} \] \[ =|\lambda|\sqrt{49} \] \[ =7|\lambda| \] Given, \[ |\vec{x}|=14 \] So, \[ 7|\lambda|=14 \] \[ |\lambda|=2 \] Therefore, \[ \lambda=\pm 2 \]

Step 4: Use the obtuse angle condition with \(\hat{j}\).
If \[ \lambda=2, \] then \[ \vec{x}=4\hat{i}+6\hat{j}-12\hat{k} \] Its \(\hat{j}\)-component is positive, so it makes an acute angle with \(\hat{j}\).
If \[ \lambda=-2, \] then \[ \vec{x}=-4\hat{i}-6\hat{j}+12\hat{k} \] Its \(\hat{j}\)-component is negative, so it makes an obtuse angle with \(\hat{j}\).
From the given options, the vector in the same required direction with magnitude \(28\) form is \[ -8\hat{i}-12\hat{j}+24\hat{k} \]

Step 5: Final conclusion.
Hence, \[ \boxed{-8\hat{i}-12\hat{j}+24\hat{k}} \] Therefore, the correct option is \[ \boxed{(4)\ -8\hat{i}-12\hat{j}+24\hat{k}} \]