Screenshot_1293.png how u are u solving plane

Answers (2)

@Santhosh

Q 90

\\Given\:\\\overrightarrow{R}\times\overrightarrow{B}=\overrightarrow{C}\times\overrightarrow{B}\\\overrightarrow{R}=\overrightarrow{C}\times\lambda\overrightarrow{B}\\Also\:given\:that\\\overrightarrow{R}\cdot \overrightarrow{A}=0\\(\overrightarrow{C}\times\lambda\overrightarrow{B})\cdot \overrightarrow{A}=0\\find\:the\:value:of\:\lambda\:you\:will\:get\:\overrightarrow{R}

@Santhosh

\\Given\:\\\overrightarrow{R}\times\overrightarrow{B}=\overrightarrow{C}\times\overrightarrow{B}\\\overrightarrow{R}=\overrightarrow{C}+\lambda\overrightarrow{B}\\Also\:given\:that\\\overrightarrow{R}\cdot \overrightarrow{A}=0\\(\overrightarrow{C}+\lambda\overrightarrow{B})\cdot \overrightarrow{A}=0\\find\:the\:value:of\:\lambda\:you\:will\:get\:\overrightarrow{R}

\\Let\:\vec R=a\hat i+b\hat j+c\hat k\\a\hat i+b\hat j+c\hat k=(5\ht i+2\hat j+5\hat k)+\lambda(6\ht i+4\hat j-8\hat k)=(5+6\lambda)\ht i+(2+4\lambda)\hat j+(5-8\lambda)\hat k\\now,\\((5+6\lambda)\ht i+(2+4\lambda)\hat j+(5-8\lambda)\hat k)\cdot(1\hat i-3\hat j+4\hat k)=0\\5+6\lambda-6-12\lambda+20-32\lambda=0\\ \lambda=1/2

\\\vec R=(5+6\lambda)\ht i+(2+4\lambda)\hat j+(5-8\lambda)\hat k\\ \vec R=(5+3)\ht i+(2+2)\hat j+(5-4)\hat k\\\vec R=(8)\ht i+(4)\hat j+\hat k

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