Three Dimensional Geometry Test - 64

Given:

In the right-angled triangle \(A B C\), hypotenuse \(A C=p\)

As we can see from the above diagram angle between vector \(\overrightarrow{B C}\) and \(\overrightarrow{B A}\) is \(90^{\circ} \). So \( \overrightarrow{B C} \cdot \overrightarrow{B A}=0 \)

\( \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}=\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{C A} \cdot \overrightarrow{C B} \)

\(\Rightarrow \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}=\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{A C} \cdot \overrightarrow{B C} \)

\(=\overrightarrow{A C} \cdot(\overrightarrow{A B}+\overrightarrow{B C})=\overrightarrow{A C} \cdot \overrightarrow{A C}=|\overrightarrow{A C}|^{2} \)

\(\Rightarrow \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}=p^{2}\)

The point that divides \(5 \hat{i}\) and \(5 \hat{j}\) in the ratio of \(k: 1\) is:

By division formula: \(\frac{mx_1+nx_2}{m+n}\)

\(\Rightarrow \frac{(5 \hat{j}) k+(5 \hat{i}) \cdot 1}{k+1} \)

\(\therefore  \vec{b}=\frac{5 \hat{i}+5 k \hat{j}}{k+1} \)

\(\vec{b}| \leq \sqrt{37} \)

\(\Rightarrow \frac{1}{k+1} \sqrt{25+25 k^{2}} \leq \sqrt{37} \)

\(\Rightarrow 5 \sqrt{1+k^{2}} \leq \sqrt{37}(k+1)\)

Squaring both sides

\(25\left(1+k^{2}\right) \leq 37\left(k^{2}+2 k+1\right)\)

\(\Rightarrow 6 k^{2}+37 k+6 \geq 0\)

\(\Rightarrow(6 k+1)(k+6) \geq 0\)

\(\mathrm{k} \in(-\infty,-6) \cup\left[-\frac{1}{6}, \infty\right)\)

Given:

\(\overrightarrow{ a }=4 \hat{ i }+6 \hat{ j }+0 \hat{k}, \quad b =0 \hat{i}+3 \hat{ j }+4 \hat{ k }|\vec{b}|=\sqrt{(3 j)^2+(4 k)^2}\)

\(=\sqrt{9+16}\left[\because i^2=1, j^2=1, k^2=1\right]\)

\(=\sqrt{25}\)

\(=5\)

\(\vec{a} \cdot \vec{b}=(4 \hat{ i }+6 \hat{ j }+0 \hat{k}) \cdot( b =0 \hat{i}+3 \hat{ j }+4 \hat{ k })\)

\(=4 \times 0+6 \times 3+0 \times 4\)

\(=18\)

Component of \(\vec{a}\) along \(\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{18}{5}\)

Component of \(\vec{a}\) along \(\vec{b}=\frac{18}{5} \hat{b}\)

\(=\frac{18}{5}\left(\frac{\vec{b}}{|\vec{b}|}\right)\left[\because \hat{b}=\left(\frac{\vec{b}}{|\vec{b}|}\right)\right]\)

\(=\frac{18}{5}\left(\frac{3 \hat{j}+4 \hat{k}}{5}\right)\)

\(=\frac{18(3 \hat{j}+4 \hat{k})}{25}\)

Given vector \(7 \hat{\imath}+4 \hat{\jmath}-3 \hat{k}, a=7, b=4\) and \(c=-3\).

The direction cosines of the vector aî \(+b \hat{\jmath}+c \hat{k}\) are given by \(\alpha=\pm \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \beta=\) \(\pm \frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}\) and \(\gamma=\pm \frac{\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}\)

\(\alpha=\pm \frac{7}{\sqrt{7^{2}+4^{2}+(-3)^{2}}}, \beta=\pm \frac{4}{\sqrt{7^{2}+4^{2}+(-3)^{2}}}\) and \(\gamma=\pm \frac{-3}{\sqrt{7^{2}+4^{2}+(-3)^{2}}}\)

\(\Rightarrow \alpha=\pm \frac{7}{\sqrt{74}}, \beta=\pm \frac{4}{\sqrt{74}}\) and \(\gamma=\frac{\mp 3}{\sqrt{74}}\)

\(\therefore(\alpha, \beta, \gamma)=\left(\frac{7}{\sqrt{74}}, \frac{4}{\sqrt{74}}, \frac{-3}{\sqrt{74}}\right)\) or \(\left(\frac{-7}{\sqrt{74}}, \frac{-4}{\sqrt{74}}, \frac{3}{\sqrt{74}}\right)\)

Given:

\(x(\hat{i}+\hat{j}+\hat{k})\) is a unit vector.

\(\therefore \sqrt{x^{2}+x^{2}+x^{2}}=1 \)

\(\Rightarrow \sqrt{3}|x|=1 \)

\(\Rightarrow|x|=\frac{1}{\sqrt{3}} \)

\(\Rightarrow x=\pm \frac{1}{\sqrt{3}}\)

Given:

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|\)

To Find: Angle between \(\vec{a}\) and \(\vec{b}\)

\(|\vec{a} \times \vec{b}|=|\vec{a} \cdot \vec{b}| \)

\(\Rightarrow|\vec{a}||\vec{b}| \sin \theta|\hat{n}|=|\vec{a}||\vec{b}| \cos \theta \)

\(\Rightarrow \sin \theta=\cos \theta \quad(\because|\hat{n}|=1) \)

\(\Rightarrow \tan \theta=1 \)

\(\therefore \theta=45^{\circ}\)

Given vector \(\hat{\imath}+2 \hat{\jmath}-\hat{k}_{,} a=1, b=2\) and \(c=-1\)

The direction cosines of the vector aî \(+b \hat{\jmath}+c \hat{k}\) are given by \(\alpha=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \beta=\) \(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}\) and \(\gamma=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}\)

\(\alpha=\frac{1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \beta=\frac{2}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}\) and \(\gamma=\frac{-1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}\)

\(\Rightarrow \alpha=\frac{1}{\sqrt{6}}, \beta=\frac{2}{\sqrt{6}}\) and \(\gamma=\frac{-1}{\sqrt{6}}\)

Given:

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

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

So,

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

\(\Rightarrow \overrightarrow{ a }+\overrightarrow{ b }=4 \hat{ i }+\hat{ j }-\hat{ k }\)

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

\(\Rightarrow \overrightarrow{ a }-\overrightarrow{ b }=-2 \hat{ i }+3 \hat{ j }-5 \hat{ k }\)

Then,

\( (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(4 \hat{ i }+\hat{ j }-\hat{ k }) \cdot(-2 \hat{ i }+3 \hat{ j }-5 \hat{ k })\)

\((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=-8+3+5=0\)

If \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0\)

Then angle between two vectors is \(90^{\circ}\).

Given:

â and \(\hat{b}\) are two unit vectors

\(\therefore|\hat{a}|=|\hat{b}|=1\)

Now, \((\hat{a}+\hat{b}) \times(\hat{a} \times \hat{b})\)

\(=\hat{a} \times(\hat{a} \times \hat{b})+\hat{b} \times(\hat{a} \times \hat{b}) \)

\(=(\hat{a} \cdot \hat{b}) \hat{a}-(\hat{a} \cdot \hat{a}) \hat{b}+(\hat{b} \cdot \hat{b}) \hat{a}-(\hat{b} \cdot \hat{a}) \hat{b} \)

\(=(\hat{a} \cdot \hat{b})(\hat{a}-\hat{b})-|\hat{a}|^{2} \hat{b}+\left.|| \hat{b}\right|^{2} \hat{a}(\because \cdot \hat{a} \cdot \hat{b}=\hat{b} \cdot \hat{a}) \)

\(=(\hat{a} \cdot \hat{b})(\hat{a}-\hat{b})-\hat{b}+\hat{a} \)

\(=(\hat{a} \cdot \hat{b})(\hat{a}-\hat{b})+(\hat{a}-\hat{b}) \)

\(=(\hat{a} \cdot \hat{b}+1)(\hat{a}-\hat{b})(\text { Let } \hat{a} \cdot \hat{b}+1=\lambda) \)

\(=\lambda(\hat{a}-\hat{b})\)

\(\therefore(\hat{a}+\hat{b}) \times(\hat{a} \times \hat{b})\) is parallel to \((\hat{a}-\hat{b})\).