Question

If $2(\vec a \times \vec c)+3(\vec b \times \vec c)=0$, where $\vec a=2\hat i-5\hat j+5\hat k$, $\vec b=\hat i-\hat j+3\hat k$ and $(\vec a-\vec b)\cdot\vec c=-97$, find $|\vec c \times \vec k|^2$.

• A. 218
• B. 207
• C. 165
• D. 210

Hint:

If $\vec p\times\vec q=0$, then the vectors are always parallel.

Answer 1

The Correct Option is A

To solve the problem, we use the given equations and vector properties. The equation provided in the problem is:

\(2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = 0\) 

We also have the vectors:

• \(\vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k}\)
• \(\vec{b} = \hat{i} - \hat{j} + 3\hat{k}\)

 

And an additional equation: \((\vec{a} - \vec{b}) \cdot \vec{c} = -97\)

First, we simplify the expression \(2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = 0\):

From the given expression, we infer:

\((2\vec{a} + 3\vec{b}) \times \vec{c} = 0\)

This implies that \(\vec{c}\) is parallel to \(2\vec{a} + 3\vec{b}\). Substitute the given vectors into this expression:

• \(2\vec{a} = 4\hat{i} - 10\hat{j} + 10\hat{k}\)
• \(3\vec{b} = 3\hat{i} - 3\hat{j} + 9\hat{k}\)

Adding these vectors:

\(2\vec{a} + 3\vec{b} = (4 + 3)\hat{i} + (-10 - 3)\hat{j} + (10 + 9)\hat{k}\)

\(\Rightarrow 7\hat{i} - 13\hat{j} + 19\hat{k}\)

Since \(\vec{c}\) is parallel to this vector, let's denote \(\vec{c} = k(7\hat{i} - 13\hat{j} + 19\hat{k})\), where \(k\) is a scalar.

Now, using the dot product condition:

\((\vec{a} - \vec{b}) \cdot \vec{c} = \left((2 - 1)\hat{i} + (-5 + 1)\hat{j} + (5 - 3)\hat{k}\right) \cdot \vec{c}\)

\(= \hat{i} - 4\hat{j} + 2\hat{k} \cdot k(7\hat{i} - 13\hat{j} + 19\hat{k})\)

This simplifies to:

\(k(7 - 4 \cdot (-13) + 2 \cdot 19) = -97\)

\(k(7 + 52 + 38) = -97\)

\(k \cdot 97 = -97\)

\(k = -1\)

Thus, \(\vec{c} = -1(7\hat{i} - 13\hat{j} + 19\hat{k}) = -7\hat{i} + 13\hat{j} - 19\hat{k}\).

Next, we need to find \(|\vec{c} \times \vec{k}|^2\) where \(\vec{k} = \hat{k}\):

\(\vec{c} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{vmatrix}\)

Calculating the cross product:

• \(\hat{i}(13 \cdot 1) - \hat{j}(-7 \cdot 1) + \hat{k}(0 \cdot 13)\)
• \(\Rightarrow 13\hat{i} + 7\hat{j}\)

Calculating the magnitude squared:

\(|\vec{c} \times \hat{k}|^2 = (13^2 + 7^2) = 169 + 49 = 218\)

Therefore, the solution is 218.