87888
If \(a, b, c\) are any thr\(\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)\) ee mutually perpendicular vectors of equal magnitude a, then \(|\mathbf{a}+\mathbf{b}+\mathbf{c}|\)
(A) : If , \(\hat{i}, \hat{j}, \hat{k}\) are the direction cosines and a, b, c are the direction ratio of line. \(2 \hat{i}+\hat{j}-2 \hat{k}\) \(a=2, b=1, c=-2\) \(\hat{i}=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{\sqrt{4+1+4}}=\frac{2}{3}\) \(\hat{j}=\frac{\underline{b}}{a^2+b^2+c^2}=\frac{1}{\sqrt{4+1+4}}=\frac{1}{3}\) \(\hat{k}=\frac{\underline{c}}{\sqrt{a^2+b^2+c^2}}=\frac{-2}{\sqrt{4+1+4}}=\frac{-2}{3}\) Hence all direction are cosine \((\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right)\)
COMEDK-2012
Vector Algebra
87873
The projections of a directed line segment on the coordinate axis are \(12,4,3\). The direction cosines of the line are
1 \(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
2 \(\frac{12}{13}, \frac{4}{13},-\frac{3}{13}\)
3 \(-\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
4 \(\frac{12}{13},-\frac{4}{13},-\frac{3}{13}\)
Explanation:
(A) : Assuming position vector of line segment ends points are \(\vec{A}\) and \(\vec{B}\) Now, given projection of line segment on the axis are \((12,4,3)\) So, \(A B=12 \hat{i}+4 \hat{j}+3 \hat{k}\) Now the direction cosine of line segment, \(=\frac{12}{\sqrt{12^2+4^2+3^2}}, \frac{4}{\sqrt{12^2+4^2+3^2}}, \frac{3}{\sqrt{12^2+4^2+3^2}}\) \(=\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
AMU-2011
Vector Algebra
87880
Angle made by the position vector of the point \((5,-4,-3)\) with the positive direction of \(X\) - axis is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{3}\)
Explanation:
(C): Let, A \((5,-4,-3)\) Direction ratio of \(\mathrm{OA}=5,-4,-3\) Direction ratio of \(\mathrm{x}-\) axis \(=1,0,0\) Angle, \(\theta=\frac{5 \cdot 1+(-4) \cdot 0+(-3) \cdot 0}{\sqrt{(5)^2+(-4)^2+(-3)^2} \sqrt{(1)^2+0^2+0^2}}\) \(=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}\) \(\therefore \theta=\frac{\pi}{4}\)
AP EAMCET-2021-19.08.2021
Vector Algebra
87887
The direction cosines of the vector \(\overrightarrow{\mathbf{a}}=-\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) are
(B) : Given that, \(\overrightarrow{\mathrm{a}} =-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{a}}| =\sqrt{(-2)^2+(1)^2+(-5)^2}\) \(=\sqrt{4+1+25}=\sqrt{30}\) Direction cosine are \((l, \mathrm{~m}, \mathrm{n})\) \(l=\frac{\mathrm{a}_1}{|\overrightarrow{\mathrm{a}}|}, \mathrm{m}=\frac{\mathrm{a}_2}{|\overrightarrow{\mathrm{a}}|}, \mathrm{n}=\frac{\mathrm{a}_3}{|\overrightarrow{\mathrm{a}}|}\) \(\therefore\) Direction cosine are -
87888
If \(a, b, c\) are any thr\(\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)\) ee mutually perpendicular vectors of equal magnitude a, then \(|\mathbf{a}+\mathbf{b}+\mathbf{c}|\)
(A) : If , \(\hat{i}, \hat{j}, \hat{k}\) are the direction cosines and a, b, c are the direction ratio of line. \(2 \hat{i}+\hat{j}-2 \hat{k}\) \(a=2, b=1, c=-2\) \(\hat{i}=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{\sqrt{4+1+4}}=\frac{2}{3}\) \(\hat{j}=\frac{\underline{b}}{a^2+b^2+c^2}=\frac{1}{\sqrt{4+1+4}}=\frac{1}{3}\) \(\hat{k}=\frac{\underline{c}}{\sqrt{a^2+b^2+c^2}}=\frac{-2}{\sqrt{4+1+4}}=\frac{-2}{3}\) Hence all direction are cosine \((\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right)\)
COMEDK-2012
Vector Algebra
87873
The projections of a directed line segment on the coordinate axis are \(12,4,3\). The direction cosines of the line are
1 \(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
2 \(\frac{12}{13}, \frac{4}{13},-\frac{3}{13}\)
3 \(-\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
4 \(\frac{12}{13},-\frac{4}{13},-\frac{3}{13}\)
Explanation:
(A) : Assuming position vector of line segment ends points are \(\vec{A}\) and \(\vec{B}\) Now, given projection of line segment on the axis are \((12,4,3)\) So, \(A B=12 \hat{i}+4 \hat{j}+3 \hat{k}\) Now the direction cosine of line segment, \(=\frac{12}{\sqrt{12^2+4^2+3^2}}, \frac{4}{\sqrt{12^2+4^2+3^2}}, \frac{3}{\sqrt{12^2+4^2+3^2}}\) \(=\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
AMU-2011
Vector Algebra
87880
Angle made by the position vector of the point \((5,-4,-3)\) with the positive direction of \(X\) - axis is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{3}\)
Explanation:
(C): Let, A \((5,-4,-3)\) Direction ratio of \(\mathrm{OA}=5,-4,-3\) Direction ratio of \(\mathrm{x}-\) axis \(=1,0,0\) Angle, \(\theta=\frac{5 \cdot 1+(-4) \cdot 0+(-3) \cdot 0}{\sqrt{(5)^2+(-4)^2+(-3)^2} \sqrt{(1)^2+0^2+0^2}}\) \(=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}\) \(\therefore \theta=\frac{\pi}{4}\)
AP EAMCET-2021-19.08.2021
Vector Algebra
87887
The direction cosines of the vector \(\overrightarrow{\mathbf{a}}=-\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) are
(B) : Given that, \(\overrightarrow{\mathrm{a}} =-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{a}}| =\sqrt{(-2)^2+(1)^2+(-5)^2}\) \(=\sqrt{4+1+25}=\sqrt{30}\) Direction cosine are \((l, \mathrm{~m}, \mathrm{n})\) \(l=\frac{\mathrm{a}_1}{|\overrightarrow{\mathrm{a}}|}, \mathrm{m}=\frac{\mathrm{a}_2}{|\overrightarrow{\mathrm{a}}|}, \mathrm{n}=\frac{\mathrm{a}_3}{|\overrightarrow{\mathrm{a}}|}\) \(\therefore\) Direction cosine are -
87888
If \(a, b, c\) are any thr\(\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)\) ee mutually perpendicular vectors of equal magnitude a, then \(|\mathbf{a}+\mathbf{b}+\mathbf{c}|\)
(A) : If , \(\hat{i}, \hat{j}, \hat{k}\) are the direction cosines and a, b, c are the direction ratio of line. \(2 \hat{i}+\hat{j}-2 \hat{k}\) \(a=2, b=1, c=-2\) \(\hat{i}=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{\sqrt{4+1+4}}=\frac{2}{3}\) \(\hat{j}=\frac{\underline{b}}{a^2+b^2+c^2}=\frac{1}{\sqrt{4+1+4}}=\frac{1}{3}\) \(\hat{k}=\frac{\underline{c}}{\sqrt{a^2+b^2+c^2}}=\frac{-2}{\sqrt{4+1+4}}=\frac{-2}{3}\) Hence all direction are cosine \((\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right)\)
COMEDK-2012
Vector Algebra
87873
The projections of a directed line segment on the coordinate axis are \(12,4,3\). The direction cosines of the line are
1 \(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
2 \(\frac{12}{13}, \frac{4}{13},-\frac{3}{13}\)
3 \(-\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
4 \(\frac{12}{13},-\frac{4}{13},-\frac{3}{13}\)
Explanation:
(A) : Assuming position vector of line segment ends points are \(\vec{A}\) and \(\vec{B}\) Now, given projection of line segment on the axis are \((12,4,3)\) So, \(A B=12 \hat{i}+4 \hat{j}+3 \hat{k}\) Now the direction cosine of line segment, \(=\frac{12}{\sqrt{12^2+4^2+3^2}}, \frac{4}{\sqrt{12^2+4^2+3^2}}, \frac{3}{\sqrt{12^2+4^2+3^2}}\) \(=\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
AMU-2011
Vector Algebra
87880
Angle made by the position vector of the point \((5,-4,-3)\) with the positive direction of \(X\) - axis is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{3}\)
Explanation:
(C): Let, A \((5,-4,-3)\) Direction ratio of \(\mathrm{OA}=5,-4,-3\) Direction ratio of \(\mathrm{x}-\) axis \(=1,0,0\) Angle, \(\theta=\frac{5 \cdot 1+(-4) \cdot 0+(-3) \cdot 0}{\sqrt{(5)^2+(-4)^2+(-3)^2} \sqrt{(1)^2+0^2+0^2}}\) \(=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}\) \(\therefore \theta=\frac{\pi}{4}\)
AP EAMCET-2021-19.08.2021
Vector Algebra
87887
The direction cosines of the vector \(\overrightarrow{\mathbf{a}}=-\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) are
(B) : Given that, \(\overrightarrow{\mathrm{a}} =-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{a}}| =\sqrt{(-2)^2+(1)^2+(-5)^2}\) \(=\sqrt{4+1+25}=\sqrt{30}\) Direction cosine are \((l, \mathrm{~m}, \mathrm{n})\) \(l=\frac{\mathrm{a}_1}{|\overrightarrow{\mathrm{a}}|}, \mathrm{m}=\frac{\mathrm{a}_2}{|\overrightarrow{\mathrm{a}}|}, \mathrm{n}=\frac{\mathrm{a}_3}{|\overrightarrow{\mathrm{a}}|}\) \(\therefore\) Direction cosine are -
87888
If \(a, b, c\) are any thr\(\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)\) ee mutually perpendicular vectors of equal magnitude a, then \(|\mathbf{a}+\mathbf{b}+\mathbf{c}|\)
(A) : If , \(\hat{i}, \hat{j}, \hat{k}\) are the direction cosines and a, b, c are the direction ratio of line. \(2 \hat{i}+\hat{j}-2 \hat{k}\) \(a=2, b=1, c=-2\) \(\hat{i}=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{\sqrt{4+1+4}}=\frac{2}{3}\) \(\hat{j}=\frac{\underline{b}}{a^2+b^2+c^2}=\frac{1}{\sqrt{4+1+4}}=\frac{1}{3}\) \(\hat{k}=\frac{\underline{c}}{\sqrt{a^2+b^2+c^2}}=\frac{-2}{\sqrt{4+1+4}}=\frac{-2}{3}\) Hence all direction are cosine \((\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right)\)
COMEDK-2012
Vector Algebra
87873
The projections of a directed line segment on the coordinate axis are \(12,4,3\). The direction cosines of the line are
1 \(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
2 \(\frac{12}{13}, \frac{4}{13},-\frac{3}{13}\)
3 \(-\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
4 \(\frac{12}{13},-\frac{4}{13},-\frac{3}{13}\)
Explanation:
(A) : Assuming position vector of line segment ends points are \(\vec{A}\) and \(\vec{B}\) Now, given projection of line segment on the axis are \((12,4,3)\) So, \(A B=12 \hat{i}+4 \hat{j}+3 \hat{k}\) Now the direction cosine of line segment, \(=\frac{12}{\sqrt{12^2+4^2+3^2}}, \frac{4}{\sqrt{12^2+4^2+3^2}}, \frac{3}{\sqrt{12^2+4^2+3^2}}\) \(=\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
AMU-2011
Vector Algebra
87880
Angle made by the position vector of the point \((5,-4,-3)\) with the positive direction of \(X\) - axis is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{3}\)
Explanation:
(C): Let, A \((5,-4,-3)\) Direction ratio of \(\mathrm{OA}=5,-4,-3\) Direction ratio of \(\mathrm{x}-\) axis \(=1,0,0\) Angle, \(\theta=\frac{5 \cdot 1+(-4) \cdot 0+(-3) \cdot 0}{\sqrt{(5)^2+(-4)^2+(-3)^2} \sqrt{(1)^2+0^2+0^2}}\) \(=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}\) \(\therefore \theta=\frac{\pi}{4}\)
AP EAMCET-2021-19.08.2021
Vector Algebra
87887
The direction cosines of the vector \(\overrightarrow{\mathbf{a}}=-\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) are
(B) : Given that, \(\overrightarrow{\mathrm{a}} =-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{a}}| =\sqrt{(-2)^2+(1)^2+(-5)^2}\) \(=\sqrt{4+1+25}=\sqrt{30}\) Direction cosine are \((l, \mathrm{~m}, \mathrm{n})\) \(l=\frac{\mathrm{a}_1}{|\overrightarrow{\mathrm{a}}|}, \mathrm{m}=\frac{\mathrm{a}_2}{|\overrightarrow{\mathrm{a}}|}, \mathrm{n}=\frac{\mathrm{a}_3}{|\overrightarrow{\mathrm{a}}|}\) \(\therefore\) Direction cosine are -
87888
If \(a, b, c\) are any thr\(\left(\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}\right)\) ee mutually perpendicular vectors of equal magnitude a, then \(|\mathbf{a}+\mathbf{b}+\mathbf{c}|\)
(A) : If , \(\hat{i}, \hat{j}, \hat{k}\) are the direction cosines and a, b, c are the direction ratio of line. \(2 \hat{i}+\hat{j}-2 \hat{k}\) \(a=2, b=1, c=-2\) \(\hat{i}=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{\sqrt{4+1+4}}=\frac{2}{3}\) \(\hat{j}=\frac{\underline{b}}{a^2+b^2+c^2}=\frac{1}{\sqrt{4+1+4}}=\frac{1}{3}\) \(\hat{k}=\frac{\underline{c}}{\sqrt{a^2+b^2+c^2}}=\frac{-2}{\sqrt{4+1+4}}=\frac{-2}{3}\) Hence all direction are cosine \((\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right)\)
COMEDK-2012
Vector Algebra
87873
The projections of a directed line segment on the coordinate axis are \(12,4,3\). The direction cosines of the line are
1 \(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
2 \(\frac{12}{13}, \frac{4}{13},-\frac{3}{13}\)
3 \(-\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
4 \(\frac{12}{13},-\frac{4}{13},-\frac{3}{13}\)
Explanation:
(A) : Assuming position vector of line segment ends points are \(\vec{A}\) and \(\vec{B}\) Now, given projection of line segment on the axis are \((12,4,3)\) So, \(A B=12 \hat{i}+4 \hat{j}+3 \hat{k}\) Now the direction cosine of line segment, \(=\frac{12}{\sqrt{12^2+4^2+3^2}}, \frac{4}{\sqrt{12^2+4^2+3^2}}, \frac{3}{\sqrt{12^2+4^2+3^2}}\) \(=\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\)
AMU-2011
Vector Algebra
87880
Angle made by the position vector of the point \((5,-4,-3)\) with the positive direction of \(X\) - axis is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{3}\)
Explanation:
(C): Let, A \((5,-4,-3)\) Direction ratio of \(\mathrm{OA}=5,-4,-3\) Direction ratio of \(\mathrm{x}-\) axis \(=1,0,0\) Angle, \(\theta=\frac{5 \cdot 1+(-4) \cdot 0+(-3) \cdot 0}{\sqrt{(5)^2+(-4)^2+(-3)^2} \sqrt{(1)^2+0^2+0^2}}\) \(=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}\) \(\therefore \theta=\frac{\pi}{4}\)
AP EAMCET-2021-19.08.2021
Vector Algebra
87887
The direction cosines of the vector \(\overrightarrow{\mathbf{a}}=-\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\) are
(B) : Given that, \(\overrightarrow{\mathrm{a}} =-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{a}}| =\sqrt{(-2)^2+(1)^2+(-5)^2}\) \(=\sqrt{4+1+25}=\sqrt{30}\) Direction cosine are \((l, \mathrm{~m}, \mathrm{n})\) \(l=\frac{\mathrm{a}_1}{|\overrightarrow{\mathrm{a}}|}, \mathrm{m}=\frac{\mathrm{a}_2}{|\overrightarrow{\mathrm{a}}|}, \mathrm{n}=\frac{\mathrm{a}_3}{|\overrightarrow{\mathrm{a}}|}\) \(\therefore\) Direction cosine are -
