3D Geometry and Vectors MCQ Questions & Answers in Geometry | Maths

$$\eqalign{ & \because {\left| {\vec a \times \vec c} \right|^2} = {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2} - {\left( {\vec a.\vec c} \right)^2} \cr & \Rightarrow {\left| { - \vec b} \right|^2} = 2{\left| {\vec c} \right|^2} - 16 \cr & \Rightarrow 3 = 2{\left| {\vec c} \right|^2} - 16 \cr & \Rightarrow {\left| {\vec c} \right|^2} = \frac{{19}}{2} \cr} $$

$$\eqalign{ & \overrightarrow {OP} .\overrightarrow {OQ} + \overrightarrow {OR} .\overrightarrow {OS} = \overrightarrow {OR} .\overrightarrow {OP} + \overrightarrow {OQ} .\overrightarrow {OS} \cr & \Rightarrow \left( {\overrightarrow {OQ} - \overrightarrow {OR} } \right).\overrightarrow {OP} - \left( {\overrightarrow {OQ} - \overrightarrow {OR} } \right).\overrightarrow {OS} = 0 \cr & \Rightarrow \left( {\overrightarrow {OQ} - \overrightarrow {OR} } \right).\left( {\overrightarrow {OP} - \overrightarrow {OS} } \right) = 0 \cr & \Rightarrow \overrightarrow {RQ} .\overrightarrow {SP} = 0 \cr & \Rightarrow RQ\, \bot \,SP.....(1) \cr & {\text{Also }}\overrightarrow {OR} .\overrightarrow {OP} + \overrightarrow {OQ} .\overrightarrow {OS} = \overrightarrow {OQ} .\overrightarrow {OR} + \overrightarrow {OP} .\overrightarrow {OS} \cr & \Rightarrow \overrightarrow {OR} .\left( {\overrightarrow {OP} - \overrightarrow {OQ} } \right) - \overrightarrow {OS} .\left( {\overrightarrow {OP} - \overrightarrow {OQ} } \right) = 0 \cr & \Rightarrow \left( {\overrightarrow {OP} - \overrightarrow {OQ} } \right).\left( {\overrightarrow {OR} - \overrightarrow {OS} } \right) = 0 \cr & \Rightarrow \overrightarrow {QP} .\overrightarrow {SR} = 0 \cr & \Rightarrow QP\, \bot \,SR.....(2) \cr} $$
From (1) and (2) $$S$$ should be the orthocenter of $$\Delta PQR.$$

$$\eqalign{ & \overrightarrow {OA} = \hat i - \hat j + 2\hat k,\,\,\overrightarrow {OB} = 2\hat i + \hat j - \hat k \cr & \therefore \,\overrightarrow {OC} = \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \cr & = \hat i + 2\hat j - 3\hat k \cr} $$

$$\eqalign{ & \therefore \,\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} \cr & = \left( {\hat i + 2\hat j - 3\hat k} \right) - \left( {2\hat i + \hat j - \hat k} \right) \cr & = - \hat i + \hat j - 2\hat k \cr} $$