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Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$.
If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to
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If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to
Solution:
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Verified Answer
The correct answer is:
11
$\begin{aligned} & \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \\ & 1=\lambda(1+\mathrm{x}+5)\end{aligned}$
$1=\lambda(x+6)$ ...(1)
$\begin{aligned}
& |\overrightarrow{\mathrm{d}}|=1 \\
& |\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1 \\
& |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 \\
& \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2\right)=1 \\
& \mathrm{x}^2+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^2 \\
& \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 \\
& 8 \mathrm{x}=8, \mathrm{x}=1 \\
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 4 & -5 \\
\mathrm{x} & 2 & 3
\end{array}\right|=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}} \\
& \left|\begin{array}{ccc}
0 & 0 & 1 \\
-2 & 9 & -4 \\
\mathrm{x}-2 & -1 & 3
\end{array}\right|=2-9(\mathrm{x}-2) \\
& =20-9 \mathrm{x} \\
& \text { at } \mathrm{x}=1 \\
& 20-9=11
\end{aligned}$
$1=\lambda(x+6)$ ...(1)
$\begin{aligned}
& |\overrightarrow{\mathrm{d}}|=1 \\
& |\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1 \\
& |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 \\
& \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2\right)=1 \\
& \mathrm{x}^2+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^2 \\
& \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 \\
& 8 \mathrm{x}=8, \mathrm{x}=1 \\
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 4 & -5 \\
\mathrm{x} & 2 & 3
\end{array}\right|=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}} \\
& \left|\begin{array}{ccc}
0 & 0 & 1 \\
-2 & 9 & -4 \\
\mathrm{x}-2 & -1 & 3
\end{array}\right|=2-9(\mathrm{x}-2) \\
& =20-9 \mathrm{x} \\
& \text { at } \mathrm{x}=1 \\
& 20-9=11
\end{aligned}$
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