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

$$\eqalign{ & {\text{We have }}\vec a \times \vec b = 39\vec k = \vec c \cr & {\text{Also }}\left| {\vec a} \right| = \sqrt {34} ,\,\left| {\vec b} \right| = \sqrt {45} ,\,\left| {\vec c} \right| = 39\,; \cr & \therefore \left| {\vec a} \right|:\left| {\vec b} \right|:\left| {\vec c} \right| = \sqrt {34} :\sqrt {45} :39 \cr} $$

Direction ratios of the lines are $$4\lambda + 1,\,4,\, - 18,$$     and $$ - 3,\,5\mu - 3,\,6.$$
They are parallel
$$\eqalign{ & \Rightarrow \frac{{4\lambda + 1}}{{ - 3}} = \frac{4}{{5\mu - 3}} = \frac{{ - 18}}{6} \cr & \Rightarrow \frac{{4\lambda + 1}}{{ - 3}} = - 3{\text{ and }}\frac{4}{{5\mu - 3}} = - 3 \cr} $$

$$\eqalign{ & \left( {\hat i - 2x\hat j - 3y\hat k} \right).\left( {\hat i + 3x\hat j + 2y\hat k} \right) = 0 \cr & 1 - 6{x^2} - 6{y^2} = 0 \cr & - 6{x^2} - 6{y^2} = - 1 \cr & {x^2} + {y^2} = \frac{1}{6} \cr & {x^2} + {y^2} = {\left( {\sqrt {\frac{1}{6}} } \right)^2} \cr} $$

Since, $$\vec a,\,\vec b$$  and $${\vec c}$$ are mutually orthogonal
$$\eqalign{ & \therefore \,\,\vec a.\vec b = 0,\,\,\vec b.\vec c = 0,\,\,\vec c.\vec a = 0 \cr & \Rightarrow 2\lambda + 4 + \mu = 0.....({\text{i}}) \cr & \Rightarrow \lambda - 1 + 2\mu = 0.....({\text{ii}}) \cr} $$
On solving (i) and (ii), we get $$\lambda = - 3,\,\,\mu = 2$$

$$\eqalign{ & \left( {\vec a + \vec b + \vec c} \right).\left[ {\left( {\vec a + \vec b} \right) \times \left( {\vec a + \vec c} \right)} \right] \cr & = \left( {\vec a + \vec b + \vec c} \right).\left[ {\vec a \times \vec a + \vec a \times \vec c + \vec b \times \vec a + \vec b \times \vec c} \right] \cr & = \left( {\vec a + \vec b + \vec c} \right).\left[ {\vec a \times \vec c + \vec b \times \vec a + \vec b \times \vec c} \right]\,\,\,\,\,\left[ {\because \vec a \times \vec a = 0} \right] \cr & = \vec a.\vec a \times \vec c + \vec a.\vec b \times \vec a + \vec a.\vec b \times \vec c + \vec b.\vec a \times \vec c + \vec b.\vec b \times \vec a + \vec b.\vec b \times \vec c + \vec c.\vec a \times \vec c + \vec c.\vec b \times \vec a + \vec c.\vec b \times \vec c \cr & = \left[ {\vec a\,\vec b\,\vec c} \right] - \left[ {\vec a\,\vec b\,\vec c} \right] - \left[ {\vec a\,\vec b\,\vec c} \right] \cr & = - \left[ {\vec a\,\vec b\,\vec c} \right] \cr} $$

\[\begin{array}{l} \overrightarrow A = \hat i + \hat j + \hat k\\ \overrightarrow B = 2\hat i + 3\hat j - \hat k\\ \overrightarrow A \times \overrightarrow B = \left| \begin{array}{l} \hat i\,\,\,\,\hat j\,\,\,\,\,\,\,\,\hat k\\ 1\,\,\,\,\,1\,\,\,\,\,\,\,\,1\\ 2\,\,\,\,3\,\, - 1 \end{array} \right|\\ = \hat i\left( { - 1 - 3} \right) - \hat j\left( { - 1 - 2} \right) + \hat k\left( {3 - 2} \right)\\ = - 4\hat i + 3\hat j + \hat k \end{array}\]
Vector of unit length orthogonal to both the vectors $$\overrightarrow A $$ and $$\overrightarrow B $$
$$\eqalign{ & = \frac{{\overrightarrow A \times \overrightarrow B }}{{\left| {\overrightarrow A \times \overrightarrow B } \right|}} \cr & = \frac{{ - 4\hat i + 3\hat j + \hat k}}{{\sqrt {16 + 9 + 1} }} \cr & = \frac{{ - 4\hat i + 3\hat j + \hat k}}{{\sqrt {26} }} \cr} $$

Given that $$\vec v = 2\hat i + \hat j - \hat k$$    and $$\vec w = \hat i + 3\hat k$$   and $$u$$ is a unit vector $$\therefore \,\,\left| {\vec u} \right| = 1$$
$$\eqalign{ & {\text{Now, }}\left[ {\vec u\,\vec v\,\vec w} \right] = \vec u.\left( {\vec v \times \vec w} \right) \cr & = \vec u.\left( {2\hat i + \hat j - \hat k} \right) \times \left( {\hat i + 3\hat k} \right) \cr & = \vec u.\left( {3\hat i - 7\hat j - \hat k} \right) \cr & = \sqrt {{3^2} + {7^2} + {1^2}} \,\cos \,\theta \cr} $$
which is maximum when $$\cos \,\theta = 1$$
$$\therefore $$ Maximum value of $$\left[ {\vec u\,\vec v\,\vec w} \right] = \sqrt {59} $$

$$\eqalign{ & {\text{Given }}{a_1}\overrightarrow {{r_1}} + {a_2}\overrightarrow {{r_2}} + ..... + {a_n}\overrightarrow {{r_n}} = 0 \cr & {\text{Now, }}\overrightarrow a + \overrightarrow {{r_1}} ' = \overrightarrow {{r_1}} {\text{ and so on}} \cr} $$

$$\eqalign{ & {\text{Hence,}} \cr & {a_1}\left( {\overrightarrow a + \overrightarrow {{r_1}} '} \right) + {a_2}\left( {\overrightarrow a + \overrightarrow {{r_2}} '} \right) + ..... + {a_n}\left( {\overrightarrow a + \overrightarrow {{r_n}} '} \right) = 0 \cr & {a_1}\overrightarrow {{r_1}} ' + {a_2}\overrightarrow {{r_2}} ' + ..... + {a_n}\overrightarrow {{r_n}} ' + \overrightarrow a \left( {{a_1} + {a_2} + ..... + {a_n}} \right) = 0 \cr & {\text{Hence,}} \cr & {a_1}\overrightarrow {{r_1}} ' + {a_2}\overrightarrow {{r_2}} ' + ..... + {a_n}\overrightarrow {{r_n}} ' = 0{\text{ if }}{a_1} + {a_2} + ..... + {a_n} = 0 \cr} $$

Three points $$A,\,B,\,C$$  are collinear if $$\overrightarrow {AB} \,\,||\,\,\overrightarrow {AC} $$
$$\eqalign{ & \overrightarrow {AB} = - 20\hat i - 11\hat j;\,\,\overrightarrow {AC} = \left( {a - 60} \right)\hat i - 55\hat j \cr & \overrightarrow {AB} \,\,||\,\,\overrightarrow {AC} \Rightarrow \frac{{a - 60}}{{ - 20}} = \frac{{ - 55}}{{ - 11}} \Rightarrow a = - 40 \cr} $$

$$\eqalign{ & {\text{Let }}P = \left( {\alpha ,\,\beta ,\,\gamma } \right). \cr & {\text{Then }}2\alpha + \beta - 3\gamma = 5.....(1) \cr & {\text{and }}\frac{{\alpha - 0}}{2} = \frac{{\beta - 1}}{2} = \frac{{\gamma + 1}}{{ - 3}}......(2) \cr & {\text{Solving (1) and (2),}}\,\alpha = \frac{1}{7},\,\beta = \frac{{15}}{{14}},\,\gamma = \frac{{ - 17}}{{14}} \cr & {\text{Let }}Q = \left( {\alpha ',\,\beta ',\,\gamma '} \right). \cr & {\text{Then }}2\alpha ' + \beta ' - 3\gamma ' = 5.....(3) \cr & {\text{and }}\frac{{\alpha ' + 2}}{2} = \frac{{\beta ' - 2}}{1} = \frac{{\gamma ' + 2}}{{ - 3}}......(4) \cr & {\text{Solving (3) and (4),}}\,\alpha ' = \frac{{ - 13}}{7},\,\beta ' = \frac{{29}}{{14}},\,\gamma ' = \frac{{ - 31}}{{14}} \cr} $$
$$\therefore $$  direction ratios of $$PQ$$  are $$\frac{1}{7} - \left( { - \frac{{13}}{7}} \right),\,\frac{{15}}{{14}} - \frac{{29}}{{14}},\,\frac{{ - 17}}{{14}} - \left( {\frac{{ - 31}}{{14}}} \right)$$         i.e., $$2,\, - 1,\,1.$$   So, direction cosines are $$\frac{2}{{\sqrt 6 }},\,\frac{{ - 1}}{{\sqrt 6 }},\,\frac{1}{{\sqrt 6 }}.$$