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A vector \vec{a} is parallel to the line of intersection of the plane determined by the vectors \mathrm{\hat{i}, \hat{i}+\hat{j}} and the plane determined by the vectors \mathrm{\hat{i}-\hat{j}, \hat{i}+\hat{k}}. The obtuse angle between \vec{a} and the vector \mathrm{ \vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}} is

Option: 1

\mathrm{\frac{3 \pi}{4}}


Option: 2

\frac{2 \pi}{3}


Option: 3

\frac{4 \pi}{5}


Option: 4

\frac{5 \pi}{6}


Answers (1)

best_answer

\mathrm{\overrightarrow{n_{1}} =i \times(i+j)=i \times i+i \times j=0+k=k} \\

\mathrm{\overrightarrow{n_{2}} =(i-j) \times(i+k)} \\

\mathrm{=i \times i+i \times k-j \times i-j \times k} \\

\mathrm{=-j+k-i} \\

\mathrm{=-i-j+k}

\mathrm{\therefore \overrightarrow{a} \text { is parallel to } \overrightarrow{n}_{1} \times \overrightarrow{n}_{2}}

\mathrm{=\left|\begin{array}{rrr} i & j & k \\ 0 & 0 & 1 \\ -1 & -1 & 1 \end{array}\right|} \\

\mathrm{=i-j}

\mathrm{Angle\; between \overrightarrow{a} and \overrightarrow{b}}

\mathrm{\cos \theta =\frac{\vec{a} \cdot \overrightarrow{b}}{| \overrightarrow{a}|| \overrightarrow{b}|}} \\

          \mathrm{=\frac{(i-j) \cdot(i-2 j+2 k)}{\sqrt{2} \cdot 3}} \\

          =\frac{1}{\sqrt{2}} \\

\mathrm{\therefore \text { obtuse angle }=\frac{3 \pi}{4} }

Hence correct option is 1

 

Posted by

Rishabh

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