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Let \vec{a}=4 \hat{\imath}+3 \hat{\jmath} and \vec{b}=3 \hat{\imath}-4 \hat{\jmath}+5 \hat{k}. If \vec{c} is a vector such that\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{\imath}+\hat{\jmath}+$ $\hat{k})=4, and projection of \vec{c} on \vec{a} is 1 , then the projection of \vec{c} on \vec{b} equals

Option: 1

\frac{1}{5}


Option: 2

\frac{5}{\sqrt{2}}


Option: 3

\frac{3}{\sqrt{2}}


Option: 4

\frac{1}{\sqrt{2}}


Answers (1)

best_answer

Let  \overrightarrow{\mathrm{c}}=\mathrm{c}_1 \hat{\mathrm{i}}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \hat{\mathrm{k}}                            \overrightarrow{\mathrm{c}} \cdot(\mathrm{i}+\mathrm{j}+\mathrm{k})=4

\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}                                                    \mathrm{c}_1+\mathrm{c}_2+\mathrm{c}_3=4\, \, \, \, \, \, \, \, .....(i)

\begin{aligned} & \left|\begin{array}{ccc} \mathrm{c}_1 & \mathrm{c}_2 & \mathrm{c}_3 \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{array}\right|=-25 \\ & \Rightarrow \mathrm{c}_1(15-0)-\mathrm{c}_2(20-0)+\mathrm{c}_3(-16-9)=-25 \\ & \Rightarrow 15 \mathrm{c}_1-20 \mathrm{c}_2-25 \mathrm{c}_3=-25 \\ & \Rightarrow 3 \mathrm{c}_1-4 \mathrm{c}_2-5 \mathrm{c}_3=-5 \ldots(2) \end{aligned}

Proj. of  \overrightarrow{\mathrm{c}} \text { on } \overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{a}}|}=1

\begin{aligned} & \Rightarrow \frac{(4 \hat{i}+3 \hat{\mathrm{j}})\left(\mathrm{c}_1 \hat{i}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \hat{\mathrm{k}}\right)}{\sqrt{16+9}}=1 \\ & \Rightarrow 4 \mathrm{c}_1+3 \mathrm{c}_2=5 \\ & \Rightarrow 4 \mathrm{c}_1=5-3 \mathrm{c}_2 \\ & \Rightarrow \mathrm{c}_1=\frac{5-3 \mathrm{c}_2}{4}\, \, \, \, \, \, ....(3) \end{aligned}

\mathrm{Eq}^{\mathrm{n}} \cdot \text { (1) \& (3) }                                          \mathrm{Eq}^{\mathrm{n}} \cdot \text { (2) \& (3) }

\frac{5-3 c_2}{4}+c_2+c_3=4 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad 3\left(\frac{5-3 c_2}{4}\right)-4 c_2-5 c_3=-5

5-3 c_2+4 c_2+4 c_3=16 \, \, \, \, \, \, \, \, \, \, \, \quad 15-9 \mathrm{c} 2-16 \mathrm{c} 2-20 c_3=-20

\begin{aligned} &\mathrm{c}_2+4 \mathrm{c}_3=11\, \, \, \, ....(4) &\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -25 c_2-20 c_3=-35\, \, \, \, ....(5) \end{aligned}

\mathrm{Eq}^{\mathrm{n}} \cdot \text { (4) \& (5) }

\mathrm{c}_2=11-4 \mathrm{c}_3 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad-25 \mathrm{c}_2-20 \mathrm{c}_3=-35

\mathrm{c}_2=11-4 \times 3 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad-25\left(11-4 \mathrm{c}_3\right)-20 \mathrm{c}_3=-35

  =11-12\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad 5\left(11-4 c_3\right)+4 c_3=7

\begin{aligned} &c_2=-1 &\begin{aligned} & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 55-20 \mathrm{c}_3+4 \mathrm{c}_3=7 \\ &\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -16 \mathrm{c}_3=-48 \end{aligned} \end{aligned}

\mathrm{c}_1=\frac{5-3 \mathrm{c}_2}{4} \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \mathrm{c}_3=3

=\frac{5-3(-1)}{4} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}

\mathrm{c}_1=2

   Projection of   \overrightarrow{\mathrm{c}} \text { on } \vec{b}=\left|\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|}\right|

                                                  

\begin{aligned} & \Rightarrow\left|\frac{(2 \hat{i}-\hat{j}+3 \hat{k}) \cdot(3 \hat{i}-4 \hat{j}+5 \hat{k})}{\sqrt{9+16+25}}\right| \\ & \Rightarrow\left|\frac{6+4+15}{5 \sqrt{2}}\right|=\frac{25}{5 \sqrt{2}}=\frac{5}{\sqrt{2}} \end{aligned}

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Rishi

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