Question ID: #455
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{d} = \vec{a} \times \vec{b}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} – 2\vec{a}|^2 = 8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{\pi}{4}$, then $|10 – 3(\vec{b} \cdot \vec{c})| + |\vec{d} \times \vec{c}|^2$ is equal to
Solution:
First, we find the vector $\vec{d}$ and its magnitude.
$$ \vec{d} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = -\hat{i} + \hat{j} $$
$$ |\vec{d}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} $$
Next, we use the given condition $|\vec{c} – 2\vec{a}|^2 = 8$ to find $|\vec{c}|$.
$$ |\vec{c}|^2 + 4|\vec{a}|^2 – 4(\vec{a} \cdot \vec{c}) = 8 $$
Substitute $|\vec{a}|^2 = 3$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$:
$$ |\vec{c}|^2 + 12 – 4|\vec{c}| = 8 $$
$$ (|\vec{c}| – 2)^2 = 0 \Rightarrow |\vec{c}| = 2 $$
Consequently, $\vec{a} \cdot \vec{c} = 2$.
Now, calculate $|\vec{d} \times \vec{c}|^2$ using the angle $\frac{\pi}{4}$:
$$ |\vec{d} \times \vec{c}| = |\vec{d}| |\vec{c}| \sin\left(\frac{\pi}{4}\right) = \sqrt{2} \cdot 2 \cdot \frac{1}{\sqrt{2}} = 2 $$
$$ \Rightarrow |\vec{d} \times \vec{c}|^2 = 4 $$
To find $\vec{b} \cdot \vec{c}$, we expand $\vec{d} \times \vec{c}$ using the definition $\vec{d} = \vec{a} \times \vec{b}$:
$$ \vec{d} \times \vec{c} = (\vec{a} \times \vec{b}) \times \vec{c} $$
Using the standard expansion formula $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} – (\vec{b} \cdot \vec{c})\vec{a}$:
$$ \vec{d} \times \vec{c} = 2\vec{b} – (\vec{b} \cdot \vec{c})\vec{a} $$
Let $x = \vec{b} \cdot \vec{c}$. Then $\vec{d} \times \vec{c} = 2\vec{b} – x\vec{a}$.
Squaring both sides:
$$ |\vec{d} \times \vec{c}|^2 = |2\vec{b} – x\vec{a}|^2 $$
$$ 4 = 4|\vec{b}|^2 + x^2|\vec{a}|^2 – 4x(\vec{a} \cdot \vec{b}) $$
Calculate the dot products:
$|\vec{b}|^2 = 4+4+1 = 9$
$\vec{a} \cdot \vec{b} = 2+2+1 = 5$
Substitute these values:
$$ 4 = 4(9) + 3x^2 – 20x $$
$$ 3x^2 – 20x + 32 = 0 $$
$$ (3x – 8)(x – 4) = 0 $$
$$ x = 4 \text{ or } x = \frac{8}{3} $$
Finally, calculate the required value $|10 – 3x| + 4$:
If $x = 4$: $|10 – 12| + 4 = 6$.
If $x = \frac{8}{3}$: $|10 – 8| + 4 = 6$.
Ans. 6
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