87758
The vectors \(3 \hat{i}+5 \hat{j}+2 \hat{k}, 2 \hat{i}-3 \hat{j}-5 \hat{k}\) and \(5 \hat{i}+2 \hat{j}-3 \hat{k}\) form the sides of
1 isosceles triangle
2 right triangle
3 scalene triangle
4 equilateral triangle
Explanation:
(D) : Given, \(\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}\) \(\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}\) \(\vec{c}=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{a}+\vec{b}=(3 \hat{i}+5 \hat{j}+2 \hat{k})+(2 \hat{i}-3 \hat{j}-5 \hat{k})=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\therefore\) The vectors \(\vec{a}, \vec{b}, \vec{c}\) are represented by three sides of a triangle. Now, \(|\overrightarrow{\mathrm{a}}|=|3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|=\sqrt{9+25+4}=\sqrt{38}\) \(|\overrightarrow{\mathrm{b}}|=|2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}|=\sqrt{4+9+25}=\sqrt{38}\) \(|\vec{c}|=|5 \hat{i}+2 \hat{j}-3 \hat{k}|=\sqrt{25+4+9}=\sqrt{38}\) \(\because|\vec{a}|=|\vec{b}|=|\vec{c}|\) \(\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) forms the sides of an equilateral triangle.
COMEDK-2020
Vector Algebra
87759
Forces 5P, 4P, P and 2P act along the sides \(A B\), \(B C, C D, D A\) of a square ABCD taken in order. Then the resultant is equal to
1 \(2 \mathrm{P}\)
2 \(\sqrt{5} \mathrm{P}\)
3 \(\sqrt{46} \mathrm{P}\)
4 \(2 \sqrt{5} \mathrm{P}\)
Explanation:
(D) : The resultant of 4 vectors in \(\mathrm{x}\), \(\mathrm{y}\) direction are \(5 \mathrm{P}, 4 \mathrm{P}, \mathrm{P}\) and \(2 \mathrm{P}\). Now resultant of \(x\) direction is, \(\mathrm{f}_1=4 \mathrm{P}-2 \mathrm{P}=2 \overrightarrow{\mathrm{P}}\) and resultant of direction \(\mathrm{f}_2=5 \mathrm{P}-\mathrm{P}=4 \overrightarrow{\mathrm{P}}\) The net resultant force is, \(\overrightarrow{\mathrm{R}}=2 \overrightarrow{\mathrm{P}}+4 \overrightarrow{\mathrm{P}}\) Angle of both vector is \(90^{\circ}\) So magnitude of resultant is, \(\mathrm{R}=\sqrt{\left(\mathrm{f}_1\right)^2+\left(\mathrm{f}_2\right)^2}=\sqrt{4 \mathrm{P}^2+16 \mathrm{P}^2}=\sqrt{20 \mathrm{P}^2}=2 \sqrt{5} \mathrm{P}\)
AMU-2006
Vector Algebra
87760
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vectors, then \(|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|\) equals
1 1
2 \(\sqrt{2}\)
3 \(\sqrt{3}\)
4 2
Explanation:
(C) : Given, \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vector. Then \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\) and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0\) Now, \((\vec{a}+\vec{b}-\vec{c})^2=\vec{a}^2+\vec{b}^2+(-\vec{c})^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=a^2+b^2+c^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=1+1+1+0-0-0\) \((\vec{a}+\vec{b}-\vec{c})^2=3\) \((\vec{a}+\vec{b}-\vec{c})=\sqrt{3}\)
87758
The vectors \(3 \hat{i}+5 \hat{j}+2 \hat{k}, 2 \hat{i}-3 \hat{j}-5 \hat{k}\) and \(5 \hat{i}+2 \hat{j}-3 \hat{k}\) form the sides of
1 isosceles triangle
2 right triangle
3 scalene triangle
4 equilateral triangle
Explanation:
(D) : Given, \(\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}\) \(\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}\) \(\vec{c}=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{a}+\vec{b}=(3 \hat{i}+5 \hat{j}+2 \hat{k})+(2 \hat{i}-3 \hat{j}-5 \hat{k})=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\therefore\) The vectors \(\vec{a}, \vec{b}, \vec{c}\) are represented by three sides of a triangle. Now, \(|\overrightarrow{\mathrm{a}}|=|3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|=\sqrt{9+25+4}=\sqrt{38}\) \(|\overrightarrow{\mathrm{b}}|=|2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}|=\sqrt{4+9+25}=\sqrt{38}\) \(|\vec{c}|=|5 \hat{i}+2 \hat{j}-3 \hat{k}|=\sqrt{25+4+9}=\sqrt{38}\) \(\because|\vec{a}|=|\vec{b}|=|\vec{c}|\) \(\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) forms the sides of an equilateral triangle.
COMEDK-2020
Vector Algebra
87759
Forces 5P, 4P, P and 2P act along the sides \(A B\), \(B C, C D, D A\) of a square ABCD taken in order. Then the resultant is equal to
1 \(2 \mathrm{P}\)
2 \(\sqrt{5} \mathrm{P}\)
3 \(\sqrt{46} \mathrm{P}\)
4 \(2 \sqrt{5} \mathrm{P}\)
Explanation:
(D) : The resultant of 4 vectors in \(\mathrm{x}\), \(\mathrm{y}\) direction are \(5 \mathrm{P}, 4 \mathrm{P}, \mathrm{P}\) and \(2 \mathrm{P}\). Now resultant of \(x\) direction is, \(\mathrm{f}_1=4 \mathrm{P}-2 \mathrm{P}=2 \overrightarrow{\mathrm{P}}\) and resultant of direction \(\mathrm{f}_2=5 \mathrm{P}-\mathrm{P}=4 \overrightarrow{\mathrm{P}}\) The net resultant force is, \(\overrightarrow{\mathrm{R}}=2 \overrightarrow{\mathrm{P}}+4 \overrightarrow{\mathrm{P}}\) Angle of both vector is \(90^{\circ}\) So magnitude of resultant is, \(\mathrm{R}=\sqrt{\left(\mathrm{f}_1\right)^2+\left(\mathrm{f}_2\right)^2}=\sqrt{4 \mathrm{P}^2+16 \mathrm{P}^2}=\sqrt{20 \mathrm{P}^2}=2 \sqrt{5} \mathrm{P}\)
AMU-2006
Vector Algebra
87760
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vectors, then \(|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|\) equals
1 1
2 \(\sqrt{2}\)
3 \(\sqrt{3}\)
4 2
Explanation:
(C) : Given, \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vector. Then \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\) and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0\) Now, \((\vec{a}+\vec{b}-\vec{c})^2=\vec{a}^2+\vec{b}^2+(-\vec{c})^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=a^2+b^2+c^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=1+1+1+0-0-0\) \((\vec{a}+\vec{b}-\vec{c})^2=3\) \((\vec{a}+\vec{b}-\vec{c})=\sqrt{3}\)
87758
The vectors \(3 \hat{i}+5 \hat{j}+2 \hat{k}, 2 \hat{i}-3 \hat{j}-5 \hat{k}\) and \(5 \hat{i}+2 \hat{j}-3 \hat{k}\) form the sides of
1 isosceles triangle
2 right triangle
3 scalene triangle
4 equilateral triangle
Explanation:
(D) : Given, \(\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}\) \(\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}\) \(\vec{c}=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{a}+\vec{b}=(3 \hat{i}+5 \hat{j}+2 \hat{k})+(2 \hat{i}-3 \hat{j}-5 \hat{k})=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\therefore\) The vectors \(\vec{a}, \vec{b}, \vec{c}\) are represented by three sides of a triangle. Now, \(|\overrightarrow{\mathrm{a}}|=|3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|=\sqrt{9+25+4}=\sqrt{38}\) \(|\overrightarrow{\mathrm{b}}|=|2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}|=\sqrt{4+9+25}=\sqrt{38}\) \(|\vec{c}|=|5 \hat{i}+2 \hat{j}-3 \hat{k}|=\sqrt{25+4+9}=\sqrt{38}\) \(\because|\vec{a}|=|\vec{b}|=|\vec{c}|\) \(\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) forms the sides of an equilateral triangle.
COMEDK-2020
Vector Algebra
87759
Forces 5P, 4P, P and 2P act along the sides \(A B\), \(B C, C D, D A\) of a square ABCD taken in order. Then the resultant is equal to
1 \(2 \mathrm{P}\)
2 \(\sqrt{5} \mathrm{P}\)
3 \(\sqrt{46} \mathrm{P}\)
4 \(2 \sqrt{5} \mathrm{P}\)
Explanation:
(D) : The resultant of 4 vectors in \(\mathrm{x}\), \(\mathrm{y}\) direction are \(5 \mathrm{P}, 4 \mathrm{P}, \mathrm{P}\) and \(2 \mathrm{P}\). Now resultant of \(x\) direction is, \(\mathrm{f}_1=4 \mathrm{P}-2 \mathrm{P}=2 \overrightarrow{\mathrm{P}}\) and resultant of direction \(\mathrm{f}_2=5 \mathrm{P}-\mathrm{P}=4 \overrightarrow{\mathrm{P}}\) The net resultant force is, \(\overrightarrow{\mathrm{R}}=2 \overrightarrow{\mathrm{P}}+4 \overrightarrow{\mathrm{P}}\) Angle of both vector is \(90^{\circ}\) So magnitude of resultant is, \(\mathrm{R}=\sqrt{\left(\mathrm{f}_1\right)^2+\left(\mathrm{f}_2\right)^2}=\sqrt{4 \mathrm{P}^2+16 \mathrm{P}^2}=\sqrt{20 \mathrm{P}^2}=2 \sqrt{5} \mathrm{P}\)
AMU-2006
Vector Algebra
87760
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vectors, then \(|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|\) equals
1 1
2 \(\sqrt{2}\)
3 \(\sqrt{3}\)
4 2
Explanation:
(C) : Given, \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vector. Then \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\) and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0\) Now, \((\vec{a}+\vec{b}-\vec{c})^2=\vec{a}^2+\vec{b}^2+(-\vec{c})^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=a^2+b^2+c^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=1+1+1+0-0-0\) \((\vec{a}+\vec{b}-\vec{c})^2=3\) \((\vec{a}+\vec{b}-\vec{c})=\sqrt{3}\)
87758
The vectors \(3 \hat{i}+5 \hat{j}+2 \hat{k}, 2 \hat{i}-3 \hat{j}-5 \hat{k}\) and \(5 \hat{i}+2 \hat{j}-3 \hat{k}\) form the sides of
1 isosceles triangle
2 right triangle
3 scalene triangle
4 equilateral triangle
Explanation:
(D) : Given, \(\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}\) \(\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}\) \(\vec{c}=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{a}+\vec{b}=(3 \hat{i}+5 \hat{j}+2 \hat{k})+(2 \hat{i}-3 \hat{j}-5 \hat{k})=5 \hat{i}+2 \hat{j}-3 \hat{k}\) \(\therefore\) The vectors \(\vec{a}, \vec{b}, \vec{c}\) are represented by three sides of a triangle. Now, \(|\overrightarrow{\mathrm{a}}|=|3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|=\sqrt{9+25+4}=\sqrt{38}\) \(|\overrightarrow{\mathrm{b}}|=|2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}|=\sqrt{4+9+25}=\sqrt{38}\) \(|\vec{c}|=|5 \hat{i}+2 \hat{j}-3 \hat{k}|=\sqrt{25+4+9}=\sqrt{38}\) \(\because|\vec{a}|=|\vec{b}|=|\vec{c}|\) \(\therefore \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) forms the sides of an equilateral triangle.
COMEDK-2020
Vector Algebra
87759
Forces 5P, 4P, P and 2P act along the sides \(A B\), \(B C, C D, D A\) of a square ABCD taken in order. Then the resultant is equal to
1 \(2 \mathrm{P}\)
2 \(\sqrt{5} \mathrm{P}\)
3 \(\sqrt{46} \mathrm{P}\)
4 \(2 \sqrt{5} \mathrm{P}\)
Explanation:
(D) : The resultant of 4 vectors in \(\mathrm{x}\), \(\mathrm{y}\) direction are \(5 \mathrm{P}, 4 \mathrm{P}, \mathrm{P}\) and \(2 \mathrm{P}\). Now resultant of \(x\) direction is, \(\mathrm{f}_1=4 \mathrm{P}-2 \mathrm{P}=2 \overrightarrow{\mathrm{P}}\) and resultant of direction \(\mathrm{f}_2=5 \mathrm{P}-\mathrm{P}=4 \overrightarrow{\mathrm{P}}\) The net resultant force is, \(\overrightarrow{\mathrm{R}}=2 \overrightarrow{\mathrm{P}}+4 \overrightarrow{\mathrm{P}}\) Angle of both vector is \(90^{\circ}\) So magnitude of resultant is, \(\mathrm{R}=\sqrt{\left(\mathrm{f}_1\right)^2+\left(\mathrm{f}_2\right)^2}=\sqrt{4 \mathrm{P}^2+16 \mathrm{P}^2}=\sqrt{20 \mathrm{P}^2}=2 \sqrt{5} \mathrm{P}\)
AMU-2006
Vector Algebra
87760
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vectors, then \(|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|\) equals
1 1
2 \(\sqrt{2}\)
3 \(\sqrt{3}\)
4 2
Explanation:
(C) : Given, \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular unit vector. Then \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\) and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0\) Now, \((\vec{a}+\vec{b}-\vec{c})^2=\vec{a}^2+\vec{b}^2+(-\vec{c})^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=a^2+b^2+c^2+2 \vec{a} \vec{b}-2 \vec{b} \vec{c}-2 \vec{c} \vec{a}\) \((\vec{a}+\vec{b}-\vec{c})^2=1+1+1+0-0-0\) \((\vec{a}+\vec{b}-\vec{c})^2=3\) \((\vec{a}+\vec{b}-\vec{c})=\sqrt{3}\)
