121092
Let the foot of the perpendicular from the point \((1,2,4)\) on the line \(\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}\) be \(P\). Then the distance of \(P\) from the plane \(3 x+4 y+\) \(12 z+23=0\)
121109
For \(\mathbf{A}(1,-2,4), \mathrm{B}(5,-1,7), \mathrm{C}(3,6,-2)\), \(D(4,5,-1)\) the projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{CD}}\) is \(\qquad\)
1 \((2 \sqrt{3},-2 \sqrt{3}, 2 \sqrt{3})\)
2 \(\frac{3}{13}(4,1,3)\)
3 \((1,-1,1)\)
4 \((2,-2,2)\)
Explanation:
D From Question - For \(A=(1,-2,4)=(\hat{i}-2 \hat{j}+4 \hat{k})\)
\(B=(5,-1,7)=(5 \hat{i}-\hat{j}+7 \hat{k})\)
\(C=(3,6,-2)=(3 \hat{i}+6 \hat{j}-2 \hat{k})\)
\(D=(4,5,-1)=(4 \hat{i}+5 \hat{j}-\hat{k})\)
Then,
\(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
\(=5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+7 \hat{\mathrm{k}}-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{AB}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}=(4,1,3)\)
and \(\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{D}}-\overrightarrow{\mathrm{C}}\)
\(=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}}-3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{CD}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}=(1,-1,1)\)
\( \vert\overrightarrow{\mathrm{CD}} \vert=\sqrt{(1)^2+(-1)^2+1^2}=\sqrt{3}\)
For projection of \(\overrightarrow{\mathrm{AB}}\) on \(\overrightarrow{\mathrm{CD}}=\left(\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{ \vert\overrightarrow{\mathrm{CD}} \vert^2}\right) \widehat{\mathrm{CD}}\)
\(=\frac{(4 \hat{i}+\hat{j}+3 \hat{k})(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}\)
\(=\frac{(4-1+3)}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})\)
\(\left.=\frac{6}{3}(\hat{i}-\hat{j}+\hat{k}) \quad \because \widehat{C D}=\frac{\overrightarrow{C D}}{ \vert\overrightarrow{C D} \vert}\right)\)
\(=2 \hat{i}-2 \hat{j}+2 \hat{k}\)
\(=(2,-2,2)\)
121092
Let the foot of the perpendicular from the point \((1,2,4)\) on the line \(\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}\) be \(P\). Then the distance of \(P\) from the plane \(3 x+4 y+\) \(12 z+23=0\)
121109
For \(\mathbf{A}(1,-2,4), \mathrm{B}(5,-1,7), \mathrm{C}(3,6,-2)\), \(D(4,5,-1)\) the projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{CD}}\) is \(\qquad\)
1 \((2 \sqrt{3},-2 \sqrt{3}, 2 \sqrt{3})\)
2 \(\frac{3}{13}(4,1,3)\)
3 \((1,-1,1)\)
4 \((2,-2,2)\)
Explanation:
D From Question - For \(A=(1,-2,4)=(\hat{i}-2 \hat{j}+4 \hat{k})\)
\(B=(5,-1,7)=(5 \hat{i}-\hat{j}+7 \hat{k})\)
\(C=(3,6,-2)=(3 \hat{i}+6 \hat{j}-2 \hat{k})\)
\(D=(4,5,-1)=(4 \hat{i}+5 \hat{j}-\hat{k})\)
Then,
\(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
\(=5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+7 \hat{\mathrm{k}}-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{AB}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}=(4,1,3)\)
and \(\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{D}}-\overrightarrow{\mathrm{C}}\)
\(=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}}-3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{CD}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}=(1,-1,1)\)
\( \vert\overrightarrow{\mathrm{CD}} \vert=\sqrt{(1)^2+(-1)^2+1^2}=\sqrt{3}\)
For projection of \(\overrightarrow{\mathrm{AB}}\) on \(\overrightarrow{\mathrm{CD}}=\left(\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{ \vert\overrightarrow{\mathrm{CD}} \vert^2}\right) \widehat{\mathrm{CD}}\)
\(=\frac{(4 \hat{i}+\hat{j}+3 \hat{k})(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}\)
\(=\frac{(4-1+3)}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})\)
\(\left.=\frac{6}{3}(\hat{i}-\hat{j}+\hat{k}) \quad \because \widehat{C D}=\frac{\overrightarrow{C D}}{ \vert\overrightarrow{C D} \vert}\right)\)
\(=2 \hat{i}-2 \hat{j}+2 \hat{k}\)
\(=(2,-2,2)\)
