Let `veca = 2hatj + 3hatj + 4hatk ,vecb= hati - 2hatj - 2hatk and vecc =-hati+4hatj +3hatk ` . If `vecd` is a vector perpendicular to both `vecb` and `vecc`, and `veca*vecd=18`, then `|veca times vecd|^2` is equal to

Options

A. 640
B. 680
C. 720
D. 760

Answer

To solve the problem step by step, we will first define the vectors and then find the required quantities.

### Step 1: Define the vectors
Given:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
\[ \vec{b} = \hat{i} - 2\hat{j} - 2\hat{k} \]
\[ \vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k} \]

### Step 2: Find the vector \(\vec{d}\) that is perpendicular to both \(\vec{b}\) and \(\vec{c}\)
To find a vector \(\vec{d}\) that is perpendicular to both \(\vec{b}\) and \(\vec{c}\), we can use the cross product:
\[ \vec{d} = \lambda (\vec{b} \times \vec{c}) \]
where \(\lambda\) is a scalar.

### Step 3: Calculate \(\vec{b} \times \vec{c}\)
Using the determinant method for the cross product:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} -2 & -2 \\ 4 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ -1 & 4 \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i}((-2)(3) - (-2)(4)) - \hat{j}((1)(3) - (-2)(-1)) + \hat{k}((1)(4) - (-2)(-1)) \]
\[ = \hat{i}(-6 + 8) - \hat{j}(3 - 2) + \hat{k}(4 - 2) \]
\[ = 2\hat{i} - 1\hat{j} + 2\hat{k} \]
Thus,
\[ \vec{b} \times \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} \]

### Step 4: Express \(\vec{d}\)
Now we can express \(\vec{d}\):
\[ \vec{d} = \lambda (2\hat{i} - \hat{j} + 2\hat{k}) \]

### Step 5: Use the condition \(\vec{a} \cdot \vec{d} = 18\)
Calculating \(\vec{a} \cdot \vec{d}\):
\[ \vec{a} \cdot \vec{d} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\lambda(2\hat{i} - \hat{j} + 2\hat{k})) \]
\[ = \lambda (2 \cdot 2 + 3 \cdot (-1) + 4 \cdot 2) \]
\[ = \lambda (4 - 3 + 8) = \lambda (9) \]
Setting this equal to 18:
\[ 9\lambda = 18 \Rightarrow \lambda = 2 \]

### Step 6: Find \(\vec{d}\)
Substituting \(\lambda\) back into \(\vec{d}\):
\[ \vec{d} = 2(2\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} - 2\hat{j} + 4\hat{k} \]

### Step 7: Calculate \(|\vec{a} \times \vec{d}|^2\)
Now, we need to calculate \(|\vec{a} \times \vec{d}|^2\):
\[ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & -2 & 4 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 3 & 4 \\ -2 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 4 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 4 & -2 \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i}(3 \cdot 4 - 4 \cdot (-2)) - \hat{j}(2 \cdot 4 - 4 \cdot 4) + \hat{k}(2 \cdot (-2) - 3 \cdot 4) \]
\[ = \hat{i}(12 + 8) - \hat{j}(8 - 16) + \hat{k}(-4 - 12) \]
\[ = 20\hat{i} + 8\hat{j} - 16\hat{k} \]

### Step 8: Calculate the magnitude squared
Now, we find \(|\vec{a} \times \vec{d}|^2\):
\[ |\vec{a} \times \vec{d}|^2 = (20)^2 + (8)^2 + (-16)^2 \]
\[ = 400 + 64 + 256 = 720 \]

### Final Answer
Thus, \(|\vec{a} \times \vec{d}|^2 = 720\).