Q.23. If , and be three vectors such that and , then is equal to ________.

Q: What is the correct answer for JEE Main Mathematics Q23?

The correct answer is 3. Sol. Using triple product identity $(\vec{a} \times \vec{c}) \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{a}$. Given $\vec{a} \times \vec{c} = \vec{b}$, so $\vec{b} \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{a}$. We have $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, so $\vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3$. Given $\vec{a} \

Q: Q.23. If , and be three vectors such that and , then is equal to ________.

The correct answer is 3. Sol. Using triple product identity $(\vec{a} \times \vec{c}) \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{a}$. Given $\vec{a} \times \vec{c} = \vec{b}$, so $\vec{b} \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{a}$. We have $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, so $\vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3$. Given $\vec{a} \

Question

Q.23. If ,  and  be three vectors such that  and , then  is equal to ________.

FirstInTest · Accepted Answer

Sol. Using triple product identity $(\vec{a} 	imes \vec{c}) 	imes \vec{a} = (\vec{a} \cdot \vec{a}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{a}$. Given $\vec{a} 	imes \vec{c} = \vec{b}$, so $\vec{b} 	imes \vec{a} = (\vec{a} \cdot \vec{a}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{a}$. We have $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, so $\vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3$. Given $\vec{a} \cdot \vec{c} = 3$. So, $\vec{b} 	imes \vec{a} = 3\vec{c} - 3\vec{a}$. This gives $3\vec{c} = 3\vec{a} + \vec{b} 	imes \vec{a}$. Calculate $\vec{b} 	imes \vec{a}$: $\vec{b} 	imes \vec{a} = (\hat{j} - \hat{k}) 	imes (\hat{i} + \hat{j} + \hat{k}) = (\hat{j} 	imes \hat{i}) + (\hat{j} 	imes \hat{j}) + (\hat{j} 	imes \hat{k}) - (\hat{k} 	imes \hat{i}) - (\hat{k} 	imes \hat{j}) - (\hat{k} 	imes \hat{k}) = -\hat{k} + 0 + \hat{i} - \hat{j} - (-\hat{i}) - 0 = -\hat{k} + \hat{i} - \hat{j} + \hat{i} = 2\hat{i} - \hat{j} - \hat{k}$. So, $3\vec{c} = 3(\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} - \hat{j} - \hat{k}) = 3\hat{i} + 3\hat{j} + 3\hat{k} + 2\hat{i} - \hat{j} - \hat{k} = 5\hat{i} + 2\hat{j} + 2\hat{k}$. Therefore, $\vec{c} = \frac{1}{3} (5\hat{i} + 2\hat{j} + 2\hat{k})$. We need to find $\vec{c} \cdot (\vec{a} - 2\vec{b})$. First, calculate $\vec{a} - 2\vec{b}$: $\vec{a} - 2\vec{b} = (\hat{i} + \hat{j} + \hat{k}) - 2(\hat{j} - \hat{k}) = \hat{i} + \hat{j} + \hat{k} - 2\hat{j} + 2\hat{k} = \hat{i} - \hat{j} + 3\hat{k}$. Now, $\vec{c} \cdot (\vec{a} - 2\vec{b}) = \frac{1}{3} (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (\hat{i} - \hat{j} + 3\hat{k}) = \frac{1}{3} (5(1) + 2(-1) + 2(3)) = \frac{1}{3} (5 - 2 + 6) = \frac{1}{3} (9) = 3$.

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions