Three Dimensional Geometry Test - 65

Given:

Vectors \(\alpha \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}, \hat{\mathrm{i}}+\hat{\mathrm{k}}\) and \(\hat{\gamma}+\gamma \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}\) lie on a plane Vectors lie on the same plane so vectors are coplanar.

Condition for coplanarity: \(\vec{a} .(\vec{b} \times \vec{c})=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=0\)

Therefore, \(\left|\begin{array}{lll}\alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta\end{array}\right|=0\)

On expanding, we get:

\( \alpha[0-\gamma]-\alpha[\beta-y]+\gamma[y-0]=0 \)

\(\Rightarrow-\alpha y-\alpha \beta+\alpha y+y^{2}=0 \)

\(\Rightarrow y^{2}=\alpha \beta\)

\(\therefore \alpha, \beta, \gamma\) are in G.P.

Given:

\(\vec{a}+\vec{b}+\vec{c}\) is collinear with \(\vec{d}\)

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

\(\Rightarrow(\vec{a}+\vec{b}+\vec{c}) \times \vec{d}+\vec{d} \times \vec{d}=0(\because \vec{d} \times \vec{d}=0) \)

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

Also given:

\(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}\) is collinear with \(\overrightarrow{\mathrm{a}}\)

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

\(\Rightarrow \vec{a} \times \vec{a}+(\vec{b}+\vec{c}+\vec{d}) \times \vec{a}=0(\because \vec{a} \times \vec{a}=0) \)

\(\Rightarrow(\vec{a}+\vec{b}+\vec{c}+\vec{d}) \times \vec{a}=0\).....(ii)

On adding (i) and (ii):

\((\vec{a}+\vec{b}+\vec{c}+\vec{d}) \times \vec{a}+(\vec{a}+\vec{b}+\vec{c}+\vec{d}) \times \vec{d}=0 \)

\(\Rightarrow(\vec{a}+\vec{b}+\vec{c}+\vec{d}) \times(\vec{a}+\vec{d})=0\).....(iii)

On subtracting (i) from (ii):

\((\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}) \times \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}) \times \overrightarrow{\mathrm{d}}=0 \)

\(\Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}) \times(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{d}})=0\)

From (iii) and (iv); it is clear that \((\vec{a}+\vec{b}+\vec{c}+\vec{d})\) is collinear to \((\vec{a}+\vec{d})\) and \((\vec{a}-\vec{d})\).

\(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} ,~ \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) be three non-zero vectors such that \(\vec{c}\) is a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).

Angle between \(\vec{a} \) and \( \vec{b}\) = \(\frac{\pi}{6}\)

We know that \(\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=[(\vec{a} \times \vec{b}) \cdot \vec{c}]^{2}\)

So, \(=\left[|\vec{a} \times \vec{b}||\vec{c}| \cos 0^{\circ}\right]^{2} \)\((\because \vec{a} \times \vec{b}\) is parellel to vector \(\vec{c}\) as \(\vec{c}\)is perpendicular to both \(\vec{a}\) and \(\vec{b}) \)

\(=\left(|\vec{a}||\vec{b}| \sin \frac{\pi}{6}\right)^{2} =|\vec{a}|^{2}|\vec{b}|^{2}\left(\frac{1}{2}\right)^{2}=\frac{1}{4}|\vec{a}|^{2}|\vec{b}|^{2}\)

We know that,\(|\vec{a}|^{2}=\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right) \)

\(|\vec{b}|^{2}=\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\)

So, \(\frac{1}{4}|\vec{a}|^{2}|\vec{b}|^{2}=\frac{1}{4}\left[\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)\right]\)

Given:

\(\vec{a}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\), and \(\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\left(\lambda^{2}-1\right) \hat{\mathrm{k}}\), be coplanar vectors.

As \(a, b, c\) are co-planar vectors \([\vec{a} \vec{b} \vec{c}]=0\)

\(\left|\begin{array}{ccc}1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \lambda^{2}-1\end{array}\right|=0\)

\(\Rightarrow \lambda^{3}-2 \lambda^{2}-9 \lambda+18=0\)

\(\Rightarrow \lambda^{2}(\lambda-2)-9(\lambda-2)=0\)

\(\Rightarrow(\lambda-3)-9(\lambda+3)(\lambda-2)=0\)

\(\Rightarrow \lambda=2,3,-3\)

So, \(\lambda=2(\) as \(\vec{a}\) is parallel to \(\vec{c}\) for \({\lambda}=\pm \mathrm{3}\) )

So, \(\vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3\end{array}\right|\)

\(\Rightarrow \vec{a} \times \vec{c}= \hat{i}(2\times3-4\times4)-\hat{j}(1\times3-2\times4)+\hat{k}(1\times4-2\times2)\)

\(\Rightarrow \hat{i}(6-16)-\hat{j}(3-8)+\hat{k}(4-4)\)

\(=-10 \hat{i}+5 \hat{j}\)

Given:

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

By writing them in the component form that is:

\(\vec{a}=(2,-1,2)\) and \(\vec{b}=(-1,1,3)\)

Then the sum of the vectors can be computed as follows:

\(\vec{a}+\vec{b}=(2-1,-1+1,2+3)\)

\(=\langle 1,0,5\rangle\)

Therefore, the sum of the vectors is \(\langle 1,0,5\rangle\).

Now the unit vector can be obtained as:

\(\|\vec{a}+\vec{b}\|=\sqrt{(1)^2+(0)^2+(5)^2}\)

\(=\sqrt{1+25}=\sqrt{26}\)

Unit Vector: \(\left(\frac{1}{\sqrt{26}}, 0, \frac{5}{\sqrt{26}}\right)\)

Given:

The vector \(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) bisects the angle between the vectors \(\vec{\mathrm{c}}\) and \(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}\).

Let \(\hat{c}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}\)

where \(x^{2}+y^{2}+z^{2}=1\)...(1)

Unit vector along \(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{\sqrt{3^{2}+4^{2}}}\)

\(=\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{5}\)

The bisectors of these two is given by:

\(\mathrm{r} =\mathrm{t}(\hat{\mathrm{a}}+\hat{\mathrm{b}}) \)

\(\mathrm{r} =\mathrm{t}\left(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{x} \hat{\mathrm{k}}+\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{5}\right) \)

\(\mathrm{r} =\frac{\mathrm{t}}{5}[(5 \mathrm{x}+3) \hat{\mathrm{i}}+(5 \mathrm{y}+4) \hat{\mathrm{j}}+5 \mathrm{z} \hat{\mathrm{k}}) \)...(2)

But the bisector is given by \(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\)...(3)

Comparing (2) and (3), we get

\(\frac{\mathrm{t}}{5}(5 \mathrm{x}+3)=-1 \)

\(\Rightarrow \mathrm{x}=-\frac{5+3 \mathrm{t}}{5 \mathrm{t}} \)

\(\frac{\mathrm{t}}{5}(5 \mathrm{y}+4)=1\)

\( \Rightarrow \mathrm{y}=\frac{5-4 \mathrm{t}}{5 \mathrm{t}} \)

\(\frac{\mathrm{t}}{5}(5 \mathrm{z})=-1 \)

\(\Rightarrow \mathrm{z}=-\frac{1}{\mathrm{t}}\)

Put all the values in equation (1), we get

\(\left(-\frac{5+3 t}{5 \mathrm{t}}\right)^{2}+\left(\frac{5-4 t}{5 \mathrm{t}}\right)^{2}+\left(-\frac{1}{\mathrm{t}}\right)^{2}=1 \)

\(\Rightarrow \frac{25+9 \mathrm{t}^{2}+30 \mathrm{t}+25+16 \mathrm{t}^{2}-40 \mathrm{t}+25}{25 \mathrm{t}^{2}}=1 \)

\(\Rightarrow 25 \mathrm{t}^{2}-10 \mathrm{t}+75=25 \mathrm{t}^{2} \)

\(\Rightarrow \mathrm{t}=7.5\)

Thus,

\(\mathrm{x}=-\frac{5+3 \times 7.5}{5 \times 7.5}=-\frac{11}{15} \)

\(\mathrm{y}=\frac{5-4 \times 7.5}{5 \times 7.5}=-\frac{10}{15} \)

\(\mathrm{z}=-\frac{1}{7.5}=-\frac{2}{15} \)

\(\hat{c}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}\)

So, \( \hat{\mathrm{c}}=\frac{1}{15}(-11 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)

Given:

\(\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \widehat{\mathrm{k}} \)

\(\vec{\beta}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-4 \widehat{\mathrm{k}}\)

Let \(\beta=\beta_{1}+\beta_{2}\)

So, \(\beta_{1}=\lambda \overrightarrow{\mathrm{a}}\) and \(\beta_{2} \cdot \overrightarrow{\mathrm{a}}=0\)

Now,

\(\beta_{1}=\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \widehat{\mathrm{k}}) \)

\(\beta_{1}=3 \lambda \hat{\mathrm{i}}+4 \lambda \hat{\mathrm{j}}+5 \lambda \widehat{\mathrm{k}}\)

Now,

\(\beta_{2}=\beta-\beta_{1} \)

\(\beta_{2}=(2-3 \lambda) \hat{i}+(1-4 \lambda) \hat{\mathrm{j}}+(-4-5 \lambda) \widehat{\mathrm{k}}\)

Since, \(\beta_{2} \cdot \overrightarrow{\mathrm{a}}=0\)

So, \(3(2-3 \lambda)+4(1-4 \lambda)+5(-4-5 \lambda)=0\)

\(\Rightarrow 6-9 \lambda+4-16 \lambda-20-25 \lambda=0\)

\(\Rightarrow -10-50 \lambda=0\)

\(\Rightarrow 50 \lambda=-10\)

\(\Rightarrow \lambda=\frac{-1}{5}\)

So, \(\beta_{1}=\frac{-1}{5}(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \widehat{\mathrm{k}})\)

\(=\frac{-3}{5} \hat{\mathrm{i}}-\frac{4}{5} \hat{\mathrm{j}}-\widehat{\mathrm{k}}\)

And,

\(\beta_{2}=\frac{13}{5} \hat{\mathrm{i}}+\frac{9}{5} \hat{\mathrm{j}}-3 \widehat{\mathrm{k}}\)

So,

\(\beta=\left(\frac{-3}{5} \hat{\mathrm{i}}-\frac{4}{5} \hat{\mathrm{j}}-\widehat{\mathrm{k}}\right)+\left(\frac{13}{5} \hat{\mathrm{i}}+\frac{9}{5} \hat{\mathrm{j}}-3 \widehat{\mathrm{k}}\right)\)

\(\beta=\beta_{1}+\beta_{2}\)

\(=\left.2\hat{\mathrm{i}}-\hat{\mathrm{j}}-4\widehat{\mathrm{k}}\right.\)

The vector triple product is defined as the cross product of one vector with the cross product of the other two. The following relationship holds:

\(a\times (b\times c) = (a.c)b - (a.b)c\)

And also written as,

\(a\times (b \times c) = b (a.c) + c (a.b)\)

Given:

\(\vec{a}=4 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\vec{b}=6 \hat{i}+7 \hat{j}+8 \hat{k}\)

If \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\) and \(\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) then \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|\)

\(\Rightarrow \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 4 \\ 6 & 7 & 8\end{array}\right|\)

\(\Rightarrow|\vec{a} \times \vec{b}|=\hat{i}(24-28)-\hat{j}(32-24)+\hat{k}(28-18)\)

\(=-4 \hat{i}-8 \hat{j}+10 \hat{k}\)

\(\Rightarrow|\vec{a} \times \vec{b}|=\sqrt{(-4)^{2}+(-8)^{2}+10^{2}}\)

\(=\sqrt{180}\)

\(=6 \sqrt{5}\)

Given:

\(\vec{a}=2 \hat{i}-6 \hat{j}-3 \hat{k}\) and \(\vec{b}=4 \hat{i}+3 \hat{j}-\hat{k}\)

\(|\vec{a}|=7,|\vec{b}|=\sqrt{26} \text { and } \vec{a} \cdot \vec{b}=-7 \)

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

Put all the given value in the above formula.

\(=\frac{-7}{7 \times \sqrt{26}}\)

\(=-\frac{1}{\sqrt{26}} \)

\(\because \sin ^{2} \theta=1-\cos ^{2} \theta\)

\(=1-\frac{1}{26}=\frac{25}{26} \)

\(\Rightarrow \sin \theta=\frac{5}{\sqrt{26}}\)