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Let veca=hat i-2 hat j+hat k, vecb=hat i...

Let `veca=hat i-2 hat j+hat k, vecb=hat i-hat j+hat k` and `vec c` is nonzero vector and `vecb xx vec c= vec b xx vec a, vec a.vec c=0`. Find `vecb.vec c`.

A

`1/2`

B

`1/3`

C

`-1/2`

D

`-1/3`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( \vec{b} \cdot \vec{c} \) given the vectors \( \vec{a} \) and \( \vec{b} \) and the conditions provided. Given: - \( \vec{a} = \hat{i} - 2\hat{j} + \hat{k} \) - \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \) - \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \) - \( \vec{a} \cdot \vec{c} = 0 \) ### Step 1: Use the property of the cross product From the equation \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \), we can take the cross product with \( \vec{a} \) on both sides: \[ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\vec{b} \times \vec{a}) \] Using the vector triple product identity \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \), we can rewrite the left-hand side: \[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{a} \times (\vec{b} \times \vec{a}) \] ### Step 2: Calculate \( \vec{a} \cdot \vec{b} \) Now, we need to calculate \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (1)(1) + (-2)(-1) + (1)(1) = 1 + 2 + 1 = 4 \] ### Step 3: Substitute known values Since \( \vec{a} \cdot \vec{c} = 0 \), we substitute this into our equation: \[ 0 \cdot \vec{b} - 4 \vec{c} = \vec{a} \times (\vec{b} \times \vec{a}) \] This simplifies to: \[ -4 \vec{c} = \vec{a} \times (\vec{b} \times \vec{a}) \] ### Step 4: Calculate \( \vec{a} \times (\vec{b} \times \vec{a}) \) To find \( \vec{a} \times (\vec{b} \times \vec{a}) \), we first calculate \( \vec{b} \times \vec{a} \): \[ \vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}((-1)(1) - (1)(-2)) - \hat{j}((1)(1) - (1)(1)) + \hat{k}((1)(-2) - (-1)(1)) \] \[ = \hat{i}(-1 + 2) - \hat{j}(1 - 1) + \hat{k}(-2 + 1) = \hat{i}(1) + 0\hat{j} - \hat{k}(1) = \hat{i} - \hat{k} \] Now, we calculate \( \vec{a} \times (\hat{i} - \hat{k}) \): \[ \vec{a} \times (\hat{i} - \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 0 & -1 \end{vmatrix} = \hat{i}((-2)(-1) - (1)(0)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(0) - (-2)(1)) \] \[ = \hat{i}(2) - \hat{j}(-1 - 1) + \hat{k}(0 + 2) = 2\hat{i} + 2\hat{j} + 2\hat{k} \] ### Step 5: Solve for \( \vec{c} \) Now substituting back, we have: \[ -4 \vec{c} = 2\hat{i} + 2\hat{j} + 2\hat{k} \] Dividing by -4: \[ \vec{c} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k} \] ### Step 6: Calculate \( \vec{b} \cdot \vec{c} \) Now we find \( \vec{b} \cdot \vec{c} \): \[ \vec{b} \cdot \vec{c} = (1)(-\frac{1}{2}) + (-1)(-\frac{1}{2}) + (1)(-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} - \frac{1}{2} = -\frac{1}{2} \] Thus, the final answer is: \[ \vec{b} \cdot \vec{c} = -\frac{1}{2} \]
To solve the problem, we need to find the value of \( \vec{b} \cdot \vec{c} \) given the vectors \( \vec{a} \) and \( \vec{b} \) and the conditions provided. Given: - \( \vec{a} = \hat{i} - 2\hat{j} + \hat{k} \) - \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \) - \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \) - \( \vec{a} \cdot \vec{c} = 0 \) ...
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