Question

Let \(\vec{a}\) = 2\(\widehat{i}\)+3\(\widehat{j}\)+4\(\widehat{k}\), \(\vec{b}\) = \(\widehat{i}\)-2\(\widehat{j}\)-2\(\widehat{k}\), \(\vec{c}\) = -\(\widehat{i}\)+4\(\widehat{j}\)+3\(\widehat{k}\) and \(\vec{d}\) is a vector perpendicular to \(\vec{b}\) and \(\vec{c}\),  \(\vec{a}\).\(\vec{d}\) = 18, then find |\(\vec{a}\)x\(\vec{d}\)|²

• A. 720
• B. 700
• C. 360
• D. 300

Answer 1

The Correct Option is A

To solve the problem, we need to find \(|\vec{a} \times \vec{d}|^2\), where \(\vec{d}\) is a vector that meets specific criteria. Let's go through the solution step-by-step:

Given:

• \(\vec{a} = 2\widehat{i} + 3\widehat{j} + 4\widehat{k}\)
• \(\vec{b} = \widehat{i} - 2\widehat{j} - 2\widehat{k}\)
• \(\vec{c} = -\widehat{i} + 4\widehat{j} + 3\widehat{k}\)
• \(\vec{d}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\)
• \(\vec{a}.\vec{d} = 18\)

Since \(\vec{d}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), it can be represented as the cross product:

\(\vec{d} = \vec{b} \times \vec{c}\)

Calculate \(\vec{b} \times \vec{c}\):

• Using the formula for the cross product: 

\[\vec{b} \times \vec{c} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \\ \end{vmatrix}\]

• The determinant is: 

\[= \widehat{i}((-2)(3) - (-2)(4)) - \widehat{j}(1 \cdot 3 - (-2)(-1)) + \widehat{k}(1 \cdot 4 - (-2)(-1))\]

• Calculating further: 

\[= \widehat{i}(-6 + 8) - \widehat{j}(3 - 2) + \widehat{k}(4 + 2) \] \[ = 2\widehat{i} - \widehat{j} + 6\widehat{k}\]

Therefore, \(\vec{d} = 2\widehat{i} - \widehat{j} + 6\widehat{k}\).

Since \(\vec{a}.\vec{d} = 18\), we have:

• Calculate the dot product: 

\[\vec{a}.\vec{d} = (2)\cdot(2) + (3)\cdot(-1) + (4)\cdot(6) \] \[ = 4 - 3 + 24 \] \[ = 25\]

• So, \(\vec{d} = k(2\widehat{i} - \widehat{j} + 6\widehat{k})\) where \(k\cdot 25 = 18 \implies k = \frac{18}{25}\)

Thus, \(\vec{d} = \frac{36}{25}\widehat{i} - \frac{18}{25}\widehat{j} + \frac{108}{25}\widehat{k}\).

Now find \(\vec{a} \times \vec{d}\) and its magnitude:

• \(\vec{a} \times \vec{d} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & 3 & 4 \\ \frac{36}{25} & -\frac{18}{25} & \frac{108}{25} \\ \end{vmatrix}\)
• \(= \widehat{i}\left(\frac{3 \cdot 108}{25} + \frac{72}{25}\right) - \widehat{j}\left(\frac{2 \cdot 108}{25} - \frac{144}{25}\right) + \widehat{k}\left(\frac{2 \cdot 18}{25} + \frac{108}{25}\right)\)
• \(= \widehat{i}\left(\frac{324 + 72}{25}\right) - \widehat{j}\left(\frac{216 - 72}{25}\right) + \widehat{k}\left(\frac{36 + 108}{25}\right)\)
• \(= \widehat{i}\left(\frac{396}{25}\right) - \widehat{j}\left(\frac{144}{25}\right) + \widehat{k}\left(\frac{144}{25}\right)\)

The magnitude squared is:

• \(|\vec{a} \times \vec{d}|^2 = \left(\frac{396}{25}\right)^2 + \left(\frac{144}{25}\right)^2 + \left(\frac{144}{25}\right)^2\)
• \(= \left(\frac{156816 + 20736 + 20736}{625}\right)\)
• \(= \frac{198288}{625}\)
• \(= 720\)

Therefore, the correct answer is 720.