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 $(\vec{a} imes \vec{b}) \cdot \vec{c}$ is…

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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 $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \ & 1=\lambda(1+\mathrm{x}+5)\end{aligned}$ $1=\lambda(x+6)$ ...(1) $\begin{aligned} & |\overrightarrow{\mathrm{d}}|=1 \ & |\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1 \ & |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 \ & \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2ight)=1 \ & \mathrm{x}^2+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^2 \ & \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 \ & 8 \mathrm{x}=8, \mathrm{x}=1 \ & \left|\begin{array}{ccc} 1 & 1 & 1 \ 2 & 4 & -5 \ \mathrm{x} & 2 & 3 \end{array}ight|=(\overrightarrow{\mathrm{a}} 	imes \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}} \ & \left|\begin{array}{ccc} 0 & 0 & 1 \ -2 & 9 & -4 \ \mathrm{x}-2 & -1 & 3 \end{array}ight|=2-9(\mathrm{x}-2) \ & =20-9 \mathrm{x} \ & 	ext { at } \mathrm{x}=1 \ & 20-9=11 \end{aligned}$

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…

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