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

We know that three vector are coplanar if their scalar triple product is zero.
\[\begin{array}{l} \Rightarrow \left| \begin{array}{l} - {\lambda ^2}\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\, - {\lambda ^2}\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,\,\,1\,\,\, - {\lambda ^2} \end{array} \right| = 0\\ {R_1} \to {R_1} + {R_2} + {R_3}\\ \Rightarrow \left| \begin{array}{l} 2 - {\lambda ^2}\,\,\,\,\,2 - {\lambda ^2}\,\,\,\,\,2 - {\lambda ^2}\\ \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\, - {\lambda ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {\lambda ^2} \end{array} \right| = 0\\ \Rightarrow \left( {2 - {\lambda ^2}} \right)\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\, - {\lambda ^2}\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\,1\,\,\, - {\lambda ^2} \end{array} \right| = 0\\ \Rightarrow \left( {2 - {\lambda ^2}} \right)\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 0\,\,\,\, - \left( {1 + {\lambda ^2}} \right)\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\, - \left( {1 + {\lambda ^2}} \right) \end{array} \right| = 0\\ \left( {{R_2} - {R_1},\,\,{R_3} - {R_1}} \right)\\ \Rightarrow \left( {2 - {\lambda ^2}} \right){\left( {1 + {\lambda ^2}} \right)^2} = 0\\ \Rightarrow \lambda = \pm \sqrt 2 \end{array}\]
$$\therefore $$ Two real solutions.

Let $$\vec c = \hat a + 2\hat b$$   and $$\vec d = 5\hat a - 4\hat b$$
Since $${\vec c}$$ and $${\vec d}$$ are perpendicular to each other
$$\eqalign{ & \therefore \,\,\,\vec c.\vec d = 0\, \cr & \Rightarrow \left( {\hat a + 2\hat b} \right).\left( {5\hat a - 4\hat b} \right) = 0 \cr & \Rightarrow 5 + 6\hat a.\hat b - 8 = 0\,\,\,\,\,\,\,\,\,\left( {\because \hat a.\hat a = 1} \right) \cr & \Rightarrow \hat a.\hat b = \frac{1}{2} \cr & \Rightarrow \theta = \frac{\pi }{3} \cr} $$

$$\eqalign{ & \because \vec u + \vec v + \vec w = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore {\left| {\vec u + \vec v + \vec w} \right|^2} = 0 \cr & \Rightarrow {\left| {\vec u} \right|^2} + \,{\left| {\vec v} \right|^2} + \,{\left| {\vec w} \right|^2} + 2\left( {\vec u.\vec v + \vec v.\vec w + \,\vec w.\vec u} \right) = 0 \cr & \Rightarrow 9 + 16 + 25 + 2\left( {\vec u.\vec v + \vec v.\vec w + \,\vec w.\vec u} \right) = 0 \cr & \Rightarrow \left( {\vec u.\vec v + \vec v.\vec w + \,\vec w.\vec u} \right) = - 25 \cr} $$

$$\eqalign{ & OQ = PQ = \lambda \,\,\,\left( {{\text{say}}} \right)\,; \cr & \overrightarrow {OP} = \overrightarrow {OQ} + \overrightarrow {QP} \,;\,\overrightarrow c = \lambda \hat a + \lambda \hat b \cr} $$
Let $${\hat a}$$ and $${\hat b}$$ be unit vectors along $$\overrightarrow a $$ and $$\overrightarrow b $$ respectively,
then, $$\hat a = \frac{1}{9}\left( {7\hat i - 4\hat j - 4\hat k} \right)$$     and $$\hat b = \frac{1}{3}\left( { - 2\hat i - \hat j + 2\hat k} \right)$$
The required vector
$$\overrightarrow c = \lambda \left( {\hat a + \hat b} \right),$$    where $$\lambda $$ is a scalar
$$\eqalign{ & \Rightarrow \overrightarrow c = \lambda \left( {\frac{1}{9}\hat i - \frac{7}{9}\hat j + \frac{2}{9}\hat k} \right) \cr & \Rightarrow {\left| {\overrightarrow c } \right|^2} = {\lambda ^2}\left( {\frac{1}{{81}} + \frac{{49}}{{81}} + \frac{4}{{81}}} \right) \cr & \Rightarrow {\left| {\overrightarrow c } \right|^2} = \frac{{54}}{{81}}{\lambda ^2} \cr & \Rightarrow {\left( {3\sqrt 6 } \right)^2} = \frac{{54}}{{81}}{\lambda ^2} \cr & \Rightarrow {\lambda ^2} = 81 \cr & \Rightarrow \lambda = \pm 9 \cr} $$
Hence, $$\overrightarrow c = \left( {\hat i - 7\hat j + 2\hat k} \right)$$

The direction ratios of the line segment joining points $$A\left( {1,\,0,\,7} \right)$$   and $$B\left( {1,\,6,\,3} \right)$$   are $$0,\,6,\,-4.$$
The direction ratios of the given line are 1, 2, 3.
Clearly $$1 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0$$
So, the given line is perpendicular to line $$AB.$$
Also , the mid point of $$A$$ and $$B$$ is (1, 3, 5) which lies on the given line.
So, the image of $$B$$ in the given line is $$A,$$  because the given line is the perpendicular bisector of line segment joining points $$A$$ and $$B.$$  But statement-2 is not a correct explanation for statement-1.

Let P.V. of

$$A,\,B$$  and $$D$$ be $$\overrightarrow 0 ,\,\overrightarrow b $$  and $$\overrightarrow d ,$$ respectively.
Then P.V. of $$C,$$
$$\overrightarrow c = \overrightarrow b + \overrightarrow d $$
Also P.V. of $${A_1} = \overrightarrow b + \frac{{\overrightarrow d }}{2}$$
and P.V. of $${B_1} = \overrightarrow d + \frac{{\overrightarrow b }}{2}$$
$$ \Rightarrow \overrightarrow {A{A_1}} + \overrightarrow {A{B_1}} = \frac{3}{2}\left( {\overrightarrow b + \overrightarrow d } \right) = \frac{3}{2}\overrightarrow {AC} $$

According to question
$$\overrightarrow a = - \hat i + \hat j + \hat k$$    and $$\overrightarrow b = \hat i - \hat j + \hat k$$
Then, \[a \times b = \left| \begin{array}{l} \,\,\,\hat i\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\hat k\\ - 1\,\,\,\,\,\,1\,\,\,\,\,\,1\\ \,\,\,1\,\, - 1\,\,\,\,1 \end{array} \right|\]
$$\eqalign{ & = \hat i\left[ {1 + 1} \right] - \hat j\left[ { - 1 - 1} \right] + \hat k\left[ {1 - 1} \right] \cr & = 2\hat i + 2\hat j + 0 \cr & = 2\left( {\hat i + \hat j} \right) \cr & {\text{and }}\left| {a \times b} \right| = \sqrt {4 + 4} = 2\sqrt 2 \cr} $$
$$\therefore $$  Required unit vector $$ = \pm \frac{{2\left( {\hat i + \hat j} \right)}}{{2\sqrt 2 }} = \pm \frac{{\hat i + \hat j}}{{\sqrt 2 }}$$

Given :
$$\eqalign{ & \vec a = 2\hat i + \hat j - 2\hat k,\,\vec b = \hat i + \hat j \cr & \Rightarrow \left| {\vec a} \right| = 3 \cr & \therefore \,\vec a \times \vec b = 2\hat i - 2\hat j + \hat k \cr & \left| {\vec a \times \vec b} \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3 \cr} $$
We have $$\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = \left| {\vec a \times \vec b} \right|\left| {\vec c} \right|\,\sin \,{30^ \circ }$$
$$\eqalign{ & \Rightarrow \left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3\left| {\vec c} \right|.\frac{1}{2} \cr & \Rightarrow 3 = 3\left| {\vec c} \right|.\frac{1}{2} \cr & \therefore \left| {\vec c} \right| = 2 \cr} $$
Now $$\left| {\vec c - \vec a} \right| = 3$$
On squaring, we get
$$\eqalign{ & \Rightarrow {c^2} + {a^2} - 2\vec c.\vec a = 9 \cr & \Rightarrow 4 + 9 - 2\vec a.\vec c = 9 \cr & \Rightarrow \vec a.\vec c = 2\,\,\,\,\,\,\,\,\,\,\left[ {\because \vec c.\vec a = \vec a.\vec c} \right] \cr} $$

\[\begin{array}{l} {\left| \begin{array}{l} {a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\ {b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\ {c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3} \end{array} \right|^2} = {\left[ {\overrightarrow A \overrightarrow B \overrightarrow C } \right]^2}\\ \Rightarrow \left| \begin{array}{l} {a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\ {b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\ {c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3} \end{array} \right| = {\left( {\left( {\overrightarrow A \times \overrightarrow B } \right).\overrightarrow C } \right)^2}\\ \Rightarrow \left| \begin{array}{l} {a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\ {b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\ {c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3} \end{array} \right| = {\left\{ {\left| {\overrightarrow A } \right|\,\left| {\overrightarrow B } \right|\sin \frac{\pi }{6}\left( {\overrightarrow C } \right).\overrightarrow C } \right\}^2}\\ \Rightarrow \left| \begin{array}{l} {a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\ {b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\ {c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3} \end{array} \right| = {\left| {\overrightarrow A } \right|^2}{\left| {\overrightarrow B } \right|^2}{\left( {\frac{1}{2}} \right)^2}{\left| {\overrightarrow C } \right|^4}\\ \Rightarrow \left| \begin{array}{l} {a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\ {b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\ {c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3} \end{array} \right| = \frac{1}{4}{\left| {\overrightarrow A } \right|^2}{\left| {\overrightarrow B } \right|^2}\\ \Rightarrow \left| \begin{array}{l} {a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\ {b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\ {c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3} \end{array} \right| = \frac{1}{4}\left( {a_1^2 + a_2^2 + a_3^2} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right) \end{array}\]

Any point on the first line is $$\left( {r,\,2r,\,3r} \right),$$   and any point on the second line is $$\left( {1 + 3r',\,2 - r',\,3 + 4r'} \right).$$      For the point of intersection, $$r = 1 + 3r',\,2r = 2 - r',\,3r = 3 + 4r'\,\,\,\, \Rightarrow r = 1{\text{ and }}r' = 0.$$
So, the point of intersection of the first two lines is $$\left( {1,\,2,\,3} \right).$$   It is on the third line. So, $$\frac{{1 + k}}{3} = \frac{{2 - 1}}{2} = \frac{{3 - 2}}{h}.$$