Three Dimensional Geometry Test - 66

Let \(A=\left(x_{1}, y_{1}, z_{1}\right)\) and \(b=\left(x_{2}, y_ 2, z_{2}\right)\) be any two points of the line \(A B\).

The direction ratio of \(A B=(x, y, z)=\left(x_{1}-x_{2}, y_{1}-y_{2}, z_{1}-z_{2}\right)\)

The direction ratios of the line joining the points \(A=(-2,3,1)\) and \(B=(1,2,\), \(-3)=[(-2-1),(3-2),(1-(-3)]=(-3,1,4)\)

The direction ratios of the line joining the points \((x, y, z)=(-3,1,-2)\)

And direction cosine \(=\left(\frac{x}{r}, \frac{y}{r}, \frac{z}{r}\right)\) Where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\)

Direction cosine of the line joining the points \((-2,3,1)\) and \((1,2,-3).\)

\(\left(\frac{\mathrm{x}}{\mathrm{r}}, \frac{\mathrm{y}}{\mathrm{r}}, \frac{\mathrm{z}}{\mathrm{r}}\right)=\left(\frac{-3}{\sqrt{26}}, \frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}\right)\)

The sum of the squares of direction cosines \(\left(\frac{-3}{\sqrt{26}}\right)^{2}+\left(\frac{1}{\sqrt{26}}\right)^{2}+\left(\frac{4}{\sqrt{26}}\right)^{2}\) \(=1\)

Given:

\(\mathrm{ABCD}\) is a trapezium with AB parallel to \(\mathrm{DC}\) and \(\mathrm{DC}=3 \mathrm{AB} . \mathrm{M}\) is the midpoint of \(\mathrm{DC} \cdot \overline{\mathrm{AB}}=\overline{\mathrm{p}}, \overline{\mathrm{BC}}=\overline{\mathrm{q}}\).

\(D C\) is parallel to \(A B\) and \(D C=3 A B\).

\(\because \overline{\mathrm{AB}}=\overline{\mathrm{p}} \therefore \overline{\mathrm{DC}}=3 \overline{\mathrm{p}}\)

\(\mathrm{M}\) is the midpoint of DC.

\(\therefore \overline{\mathrm{DM}}=\overline{\mathrm{MC}}=\frac{1}{2} \overline{\mathrm{DC}}=\frac{3}{2} \overline{\mathrm{p}}\)

\(\overline{\mathrm{MB}}=\overline{\mathrm{MC}}+\overline{\mathrm{CB}}=\overline{\mathrm{MC}}-\overline{\mathrm{BC}}\)

\(=\frac{3}{2} \overline{\mathrm{p}}-\overline{\mathrm{q}}\)

\(\overrightarrow{a}=2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\overrightarrow{b}=3 \hat{i}-2 \hat{j}-\hat{k}\)

\(\vec{a} \cdot \vec{b}=(2 \hat{i}+\hat{j}-3 \hat{k})(3 \hat{i}-2 \hat{j}-\hat{k}) \)

\(=6-2+3=7 \ldots .(1) \)

\(|\vec{a}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{14} \ldots(2) \)

\(|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+(-1)^{2}}=\sqrt{14} \ldots .(3) \)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\)

Put the values from (1), (2) and (3) in above equation:

\(=\frac{7}{\sqrt{14}. \sqrt{14}}=\frac{1}{2} \)

\(\cos \theta=\cos 60^{\circ} , \theta=60^{\circ}\)

Given:

\(\vec{A}=\hat{i}+6 i+6 k, \vec{B}=-4 \hat{i}+9 \hat{i}+6 \hat{k}, \vec{G}=\frac{-5}{3} \hat{i}+\frac{22}{3} \hat{j}+\frac{22}{3} \hat{k} .\)

Centroid of a Triangle is given by \(\mathrm{G}=\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}}{3}\).

Putting values in above equation:

\(\frac{-5 \hat{\mathrm{i}}+15 \hat{\mathrm{j}}+12 \mathrm{k}+\mathrm{C}}{3}=\frac{-5 \hat{\mathrm{i}}}{3}+\frac{22}{3} \hat{\mathrm{j}}+\frac{22}{3} \hat{\mathrm{k}} \)

\(\overrightarrow{\mathrm{C}}=7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}} \)   

\(AB = AC\) (isosceles triangle)

Given vectors are \(a \vec{i}+\vec{j}+\vec{k}, \vec{i}+b \vec{j}+\vec{k}, \vec{i}+\vec{j}+c \vec{k}~(a, b, c\) are not equal to 1\()\)

For coplanar \(\left|\begin{array}{ccc}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0\)

Applying \(c_{1} \rightarrow c_{1}-c_{2}, \quad c_{2} \rightarrow c_{2}-c_{3}\)

\(\left|\begin{array}{ccc}a-1 & 0 & 1 \\ 1-b & b-1 & 1 \\ 0 & 1-c & c\end{array}\right|=0\)

\(\Rightarrow c(a-1)(b-1)+(1-b)(1-c)-(1-c)(a-1)=0\)

\(\Rightarrow \frac{c}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=0\)

\(\Rightarrow \frac{c}{1-c}+1+\frac{1}{1-a}+\frac{1}{1-b}=1\)

\(\Rightarrow \frac{1}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=1\)

Given:

\(\overrightarrow{b} =\hat{ i} +\hat{ j}\) and \(c = \hat{j }+ \hat{k}\)

The equation of bisector of \(\overrightarrow{b}\) and \(\overrightarrow{c}\) is:

\(r=\lambda(\overrightarrow{b}+\overrightarrow{c})\)

\(=\lambda\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}+\frac{\hat{j}+\hat{k}}{\sqrt{2}}\right)\)

\(=\frac{\lambda}{\sqrt{2}}(\hat{i}+2\hat{ j}+\hat{k}) \ldots . .( i )\)

Since, vector a lies in plane of \(\overrightarrow{b}\) and \(\overrightarrow{c}\).

\(9\therefore \overrightarrow{a} = \overrightarrow{b }+\overrightarrow{c} \)

Put the given values in above equation:

\(\frac{\lambda}{\sqrt{2}}( \hat{i }+2 \hat{j} + \hat{k} )=( \hat{i} + \hat{j} )+\mu(\hat{ j} + \hat{k} )\)

On equating the coefficient of \(\hat{i}\) both sides, we get:

\(\frac{\lambda}{\sqrt{2}}=1\)

\(\Rightarrow \lambda=\sqrt{2}\)

On putting \(\lambda=\sqrt{2}\) in Eq. (i), we get:

\(r=\hat{i}+2\hat{ j}+\hat{k}\)

Since the given vector a represents the same bisector equation \(r\).

\(\therefore \alpha=1\) and \(\beta=1\)

\(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of the points \(A, B, C\) respectively and \(2 \vec{a}+\) \(3 \vec{\mathrm{b}}-5 \vec{\mathrm{c}}={0}\).

Let the ratio be \(\lambda: 1\).

Position vector of point (C): \(\vec{c}=\frac{\lambda \vec{a}+\vec{b}}{\lambda+1}\)

Put the value of \(\vec{c}\) in \( 2 \vec{a}+3 \vec{b}-5 \vec{c}=0\)

\(2 \vec{a}+3 \vec{b}-5 \times\left(\frac{\lambda \vec{a}+\vec{b}}{\lambda+1}\right)=0 \)

\(\Rightarrow \frac{(\lambda+1)(2 \vec{a}+3 \vec{b})-5 \lambda \vec{a}-5 \vec{b}}{(\lambda+1)}=0 \)

\(\Rightarrow 2 \vec{a} \lambda+3 \vec{b} \lambda+2 \vec{a}+3 \vec{b}-5 \lambda \vec{a}-5 \vec{b}=0 \)

\(\Rightarrow 3 \vec{b} \lambda-3 \vec{a} \lambda+2 \vec{a}-2 \vec{b}=0 \)

\(\Rightarrow 3 \lambda(\vec{b}-\vec{a})=2 \vec{b}-2 \vec{a} \)

\(\Rightarrow 3 \lambda=\frac{2(\vec{b}-\vec{a})}{(\vec{b}-\vec{a})}\)

\(\Rightarrow \frac{\lambda}{1}=\frac{3}{2}\)

Point \(\mathrm{C}\) divides line segment \(\mathrm{AB}\) in the ratio \(3: 2 .\)

\(c=x a+y b+z(a \times b)\)

Let \(c \cdot a = x , c \cdot b = y , x = y =\cos \theta\)

Now, C.C \(=|c|^2\)

So, \([ xa + yb + z ( a \times b )] \cdot[ xa + yb + z ( a \times b )]=| c |^2\)

\(\Rightarrow 2 x ^2+ z ^2| a \times b |^2=1[\because c\) is a unit vector\(]\)

\(\Rightarrow 2 x ^2+ z ^2\left[| a |^2| b |^2-( a \cdot b )^2\right]=1\)

\(\Rightarrow 2 x ^2+ z ^2[1-0]=1 \quad[\because a \perp b \therefore a \cdot b =0]\)

\(\Rightarrow 2 x ^2+ z ^2=1\Rightarrow z ^2=1-2 x ^2\)

\(=1-2 \cos ^2 \theta=-\cos 2 \theta\)

Given:

\(\vec{a}=\hat{j}-\hat{k}\) and \(\vec{c}=\hat{i}-\hat{j}-\hat{k}\)

Consider, \((\overrightarrow{ a } \times \overrightarrow{ b })+\overrightarrow{ c }=0\)

Multiply by \(\vec{a}\) on both side, we have

\(\Rightarrow \vec{a} \times[(\vec{a} \times \vec{b})+\vec{c}]=0\)

\(\Rightarrow \vec{a} \times[(\vec{a} \times \vec{b})]+(\vec{a} \times \vec{c})=0\)

\(\Rightarrow(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}+(\vec{a} \times \vec{c})=0 \ldots . .( i )\)

Given \(\overrightarrow{ a } \cdot \overrightarrow{ b }=3, \overrightarrow{ a } \cdot \overrightarrow{ a }=2\) and \(\overrightarrow{ a } \times \overrightarrow{ b }=-2 \overrightarrow{ i }-\overrightarrow{ j }-\overrightarrow{ k }\)

Equation (i) becomes,

\(\Rightarrow 3 \vec{a}-2 \vec{b}+(-2 \vec{i}-\vec{j}-\vec{k})=0\)

\(\Rightarrow 2 \vec{b}=3 \vec{a}+(-2 \vec{i}-\vec{j}-\vec{k})\)

\(\Rightarrow 2 \vec{b}=3(\vec{j}-\vec{k})+(-2 \vec{i}-\vec{j}-\vec{k})\)

\(\Rightarrow \vec{b}=-\hat{i}+\hat{j}-2 \hat{k}\)

So, if \(\overrightarrow{ a }=\hat{ j }-\hat{ k }\) and \(\overrightarrow{ c }=\hat{ i }-\hat{ j }-\hat{ k }\). Then the vector \(\overrightarrow{ b }\) satisfying \((\vec{a} \times \vec{b})+\vec{c}=0\) and \(\vec{a} \cdot \vec{b}=3\), is \(\vec{b}=-\hat{i}+\hat{j}-2 \hat{k}\).

Given:

The position vector of point \(\mathrm{P}=2 \hat{i}+3 \hat{j}+4 \hat{k}\)

Position Vector of point \(Q=4 \hat{i}+\hat{j}-2 \hat{k}\)

As we know,

The position vector of \(A\left(x_{1}, y_{1}, z_{1}\right)\) is given by \(\overrightarrow{O A}=x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k}\) Compute the position vector of the midpoint of \(\mathrm{P}\) and \(\mathrm{Q}\) using the formula:

If \(C(x, y, z)\) is the midpoint of segment \(\overrightarrow{A B}\), then we have the position vector of \(C\) \(=\overrightarrow{O C}=\frac{\overrightarrow{O A}+\overrightarrow{O B}}{2}\)

\(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)

The position vector of \(R\) which divides \(P Q\) in half is given by:

\(\vec{r}=\frac{2 \hat{i}+3 \hat{j}+4 \hat{k}+4 \hat{i}+\hat{j}-2 \hat{k}}{2} \)

\(\vec{r}=\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}\)

\(=3 \hat{i}+2 \hat{j}+\hat{k}\)