87676
If the position vectors of the vertices \(A, B\) and \(C\) are \(6 \hat{i}, 6 \hat{j}\) and \(\hat{k}\) respectively with respect to origin \(O\), the volume of the tetrahedron \(O A B C\) is
1 6
2 3
3 \(\frac{1}{6}\)
4 \(\frac{1}{3}\)
Explanation:
(A) : Given, \(\overrightarrow{\mathrm{OA}}=6 \hat{\mathrm{i}}=6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OB}}=6 \hat{\mathrm{j}}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{k}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) The position vectors of \(A, B, C\), with respect to origin respectively. \(\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\left|\begin{array}{lll} 6 &0& 0\\ 0& 6& 0\\ 0& 0& 1 \end{array}\right|=6(6-0)-0+0=36\) Now, volume of tetrahedron \(=\frac{1}{6}\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\frac{1}{6}(36)=6\)
MHT CET-2012
Vector Algebra
87677
If \(A(x, 2,8), B(3, y, 4)\) and \(C(4,1, z)\) are vertices of \(\triangle \mathrm{ABC}\) and \(\mathrm{G}(2,1,5)\) is the centroid then the values of \(x, y\) and \(z\) are respectively
1 \((1,0,2)\)
2 \((-1,0,2)\)
3 \((1,0,3)\)
4 \((-1,0,3)\)
Explanation:
(D) : Let, \(\vec{a}, \vec{b}, \vec{c}, \vec{g}\) be the position vectors of \(A, B, C\) and \(G\) respectively. \(\therefore \quad \vec{a}=x \hat{i}+2 \hat{j}+8 \hat{k}\) \(\vec{b}=3 \hat{i}+y \hat{j}+4 \hat{k}\) \(\vec{c}=4 \hat{i}+\hat{j}+z \hat{k}\) \(\vec{g}=2 \hat{i}+\hat{j}+5 \hat{k}\) \(A s G\) is the centroid of \(\triangle A B C, \vec{g}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\) i.e. \(3 \overrightarrow{\mathrm{g}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\) \(3(2 \hat{i}+\hat{j}+5 \hat{k})=(x \hat{i}+2 \hat{j}+8 \hat{k})+(3 \hat{i}+y \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}+z \hat{k})\) \(6 \hat{i}+3 \hat{j}+15 \hat{k}=(x+3+4) \hat{i}+(2+y+1) \hat{j}+(8+4+z) \hat{k}\) \(=(\mathrm{x}+7) \hat{\mathrm{i}}+(\mathrm{y}+3) \hat{\mathrm{j}}+(\mathrm{z}+12) \hat{\mathrm{k}}\) \(x+7=6\) i.e. \(x=-1, y+3=3\) i.e. \(y=0\) and \(z+12=15\) i.e. \(z=3\) So, \((x, y, z)=(-1,0,3)\)
MHT CET-2010
Vector Algebra
87678
If \(2 \vec{a}+3 \vec{b}-5 \vec{c}=0\), then the ratio in which \(C\) divided \(A B\) is
1 \(2: 3\) internally
2 \(2: 3\) externally
3 \(3: 2\) internally
4 \(3: 2\) externally
Explanation:
(C) : Given, \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-5 \overrightarrow{\mathrm{c}}=0\) \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}=5 \overrightarrow{\mathrm{c}}\) \(\frac{2 \vec{a}+3 \vec{b}}{5}=\vec{c}, \quad \frac{2 \vec{a}+3 \vec{b}}{3+2}=\vec{c}\) \(\frac{2\left(\overrightarrow{\mathrm{a}}+\frac{3}{2} \overrightarrow{\mathrm{b}}\right)}{2\left(\frac{3}{2}+1\right)}=\overrightarrow{\mathrm{c}} \text {, }\) \(\frac{\left(\vec{a}+\frac{3}{2} \vec{b}\right)}{\left(1+\frac{3}{2}\right)}=\vec{c}\) We know that, \(c=\frac{a+\lambda b}{1+\lambda}\) From comparing equation (i) \& (ii), we get - \(\lambda=\frac{3}{2}\) Hence, the point \(C\) divides \(A B\) internally in the ratio of \(3: 2\).
MHT CET-2009
Vector Algebra
87681
If the position vectors of the vertices \(A, B, C\) of a triangle \(\mathrm{ABC}\) are \(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}+6 \hat{\mathbf{j}}+6 \hat{\mathrm{k}}\) and \(-4 \hat{i}+9 \hat{j}+6 \hat{k}\) respectively, the triangle is :
1 equilateral
2 isosceles
3 scalene
4 right angled and isosceles also
Explanation:
(D) : Given, The position vectors of the vertices, \(\overrightarrow{\mathrm{OA}}=7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}},\) \(\overrightarrow{\mathrm{OC}}=-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\) \(\overrightarrow{\mathrm{AB}}=(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\sqrt{1+1+16}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}\) \(\overrightarrow{\mathrm{BC}}=(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{BC}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(\mid \overrightarrow{\mathrm{BC}}=\sqrt{(-3)^2+(3)^2}\) \(\overrightarrow{\mathrm{BC}}=\sqrt{18}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}\) \(\overrightarrow{\mathrm{CA}}=(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})-(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{CA}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{(4)^2+(-2)^2+(4)^2}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{36}=6\) \(3 \sqrt{2}, 3 \sqrt{2}\) \& 6 are sides of a right angled \(\Delta\). \(|\mathrm{AB}|^2+|\mathrm{BC}|^2=|\mathrm{AC}|^2\) \(\because(3 \sqrt{2})^2+(3 \sqrt{2})^2=36\) Hence, the \(\triangle A B C\) is a right-angled and isosceles also.
87676
If the position vectors of the vertices \(A, B\) and \(C\) are \(6 \hat{i}, 6 \hat{j}\) and \(\hat{k}\) respectively with respect to origin \(O\), the volume of the tetrahedron \(O A B C\) is
1 6
2 3
3 \(\frac{1}{6}\)
4 \(\frac{1}{3}\)
Explanation:
(A) : Given, \(\overrightarrow{\mathrm{OA}}=6 \hat{\mathrm{i}}=6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OB}}=6 \hat{\mathrm{j}}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{k}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) The position vectors of \(A, B, C\), with respect to origin respectively. \(\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\left|\begin{array}{lll} 6 &0& 0\\ 0& 6& 0\\ 0& 0& 1 \end{array}\right|=6(6-0)-0+0=36\) Now, volume of tetrahedron \(=\frac{1}{6}\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\frac{1}{6}(36)=6\)
MHT CET-2012
Vector Algebra
87677
If \(A(x, 2,8), B(3, y, 4)\) and \(C(4,1, z)\) are vertices of \(\triangle \mathrm{ABC}\) and \(\mathrm{G}(2,1,5)\) is the centroid then the values of \(x, y\) and \(z\) are respectively
1 \((1,0,2)\)
2 \((-1,0,2)\)
3 \((1,0,3)\)
4 \((-1,0,3)\)
Explanation:
(D) : Let, \(\vec{a}, \vec{b}, \vec{c}, \vec{g}\) be the position vectors of \(A, B, C\) and \(G\) respectively. \(\therefore \quad \vec{a}=x \hat{i}+2 \hat{j}+8 \hat{k}\) \(\vec{b}=3 \hat{i}+y \hat{j}+4 \hat{k}\) \(\vec{c}=4 \hat{i}+\hat{j}+z \hat{k}\) \(\vec{g}=2 \hat{i}+\hat{j}+5 \hat{k}\) \(A s G\) is the centroid of \(\triangle A B C, \vec{g}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\) i.e. \(3 \overrightarrow{\mathrm{g}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\) \(3(2 \hat{i}+\hat{j}+5 \hat{k})=(x \hat{i}+2 \hat{j}+8 \hat{k})+(3 \hat{i}+y \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}+z \hat{k})\) \(6 \hat{i}+3 \hat{j}+15 \hat{k}=(x+3+4) \hat{i}+(2+y+1) \hat{j}+(8+4+z) \hat{k}\) \(=(\mathrm{x}+7) \hat{\mathrm{i}}+(\mathrm{y}+3) \hat{\mathrm{j}}+(\mathrm{z}+12) \hat{\mathrm{k}}\) \(x+7=6\) i.e. \(x=-1, y+3=3\) i.e. \(y=0\) and \(z+12=15\) i.e. \(z=3\) So, \((x, y, z)=(-1,0,3)\)
MHT CET-2010
Vector Algebra
87678
If \(2 \vec{a}+3 \vec{b}-5 \vec{c}=0\), then the ratio in which \(C\) divided \(A B\) is
1 \(2: 3\) internally
2 \(2: 3\) externally
3 \(3: 2\) internally
4 \(3: 2\) externally
Explanation:
(C) : Given, \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-5 \overrightarrow{\mathrm{c}}=0\) \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}=5 \overrightarrow{\mathrm{c}}\) \(\frac{2 \vec{a}+3 \vec{b}}{5}=\vec{c}, \quad \frac{2 \vec{a}+3 \vec{b}}{3+2}=\vec{c}\) \(\frac{2\left(\overrightarrow{\mathrm{a}}+\frac{3}{2} \overrightarrow{\mathrm{b}}\right)}{2\left(\frac{3}{2}+1\right)}=\overrightarrow{\mathrm{c}} \text {, }\) \(\frac{\left(\vec{a}+\frac{3}{2} \vec{b}\right)}{\left(1+\frac{3}{2}\right)}=\vec{c}\) We know that, \(c=\frac{a+\lambda b}{1+\lambda}\) From comparing equation (i) \& (ii), we get - \(\lambda=\frac{3}{2}\) Hence, the point \(C\) divides \(A B\) internally in the ratio of \(3: 2\).
MHT CET-2009
Vector Algebra
87681
If the position vectors of the vertices \(A, B, C\) of a triangle \(\mathrm{ABC}\) are \(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}+6 \hat{\mathbf{j}}+6 \hat{\mathrm{k}}\) and \(-4 \hat{i}+9 \hat{j}+6 \hat{k}\) respectively, the triangle is :
1 equilateral
2 isosceles
3 scalene
4 right angled and isosceles also
Explanation:
(D) : Given, The position vectors of the vertices, \(\overrightarrow{\mathrm{OA}}=7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}},\) \(\overrightarrow{\mathrm{OC}}=-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\) \(\overrightarrow{\mathrm{AB}}=(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\sqrt{1+1+16}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}\) \(\overrightarrow{\mathrm{BC}}=(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{BC}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(\mid \overrightarrow{\mathrm{BC}}=\sqrt{(-3)^2+(3)^2}\) \(\overrightarrow{\mathrm{BC}}=\sqrt{18}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}\) \(\overrightarrow{\mathrm{CA}}=(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})-(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{CA}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{(4)^2+(-2)^2+(4)^2}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{36}=6\) \(3 \sqrt{2}, 3 \sqrt{2}\) \& 6 are sides of a right angled \(\Delta\). \(|\mathrm{AB}|^2+|\mathrm{BC}|^2=|\mathrm{AC}|^2\) \(\because(3 \sqrt{2})^2+(3 \sqrt{2})^2=36\) Hence, the \(\triangle A B C\) is a right-angled and isosceles also.
87676
If the position vectors of the vertices \(A, B\) and \(C\) are \(6 \hat{i}, 6 \hat{j}\) and \(\hat{k}\) respectively with respect to origin \(O\), the volume of the tetrahedron \(O A B C\) is
1 6
2 3
3 \(\frac{1}{6}\)
4 \(\frac{1}{3}\)
Explanation:
(A) : Given, \(\overrightarrow{\mathrm{OA}}=6 \hat{\mathrm{i}}=6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OB}}=6 \hat{\mathrm{j}}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{k}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) The position vectors of \(A, B, C\), with respect to origin respectively. \(\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\left|\begin{array}{lll} 6 &0& 0\\ 0& 6& 0\\ 0& 0& 1 \end{array}\right|=6(6-0)-0+0=36\) Now, volume of tetrahedron \(=\frac{1}{6}\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\frac{1}{6}(36)=6\)
MHT CET-2012
Vector Algebra
87677
If \(A(x, 2,8), B(3, y, 4)\) and \(C(4,1, z)\) are vertices of \(\triangle \mathrm{ABC}\) and \(\mathrm{G}(2,1,5)\) is the centroid then the values of \(x, y\) and \(z\) are respectively
1 \((1,0,2)\)
2 \((-1,0,2)\)
3 \((1,0,3)\)
4 \((-1,0,3)\)
Explanation:
(D) : Let, \(\vec{a}, \vec{b}, \vec{c}, \vec{g}\) be the position vectors of \(A, B, C\) and \(G\) respectively. \(\therefore \quad \vec{a}=x \hat{i}+2 \hat{j}+8 \hat{k}\) \(\vec{b}=3 \hat{i}+y \hat{j}+4 \hat{k}\) \(\vec{c}=4 \hat{i}+\hat{j}+z \hat{k}\) \(\vec{g}=2 \hat{i}+\hat{j}+5 \hat{k}\) \(A s G\) is the centroid of \(\triangle A B C, \vec{g}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\) i.e. \(3 \overrightarrow{\mathrm{g}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\) \(3(2 \hat{i}+\hat{j}+5 \hat{k})=(x \hat{i}+2 \hat{j}+8 \hat{k})+(3 \hat{i}+y \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}+z \hat{k})\) \(6 \hat{i}+3 \hat{j}+15 \hat{k}=(x+3+4) \hat{i}+(2+y+1) \hat{j}+(8+4+z) \hat{k}\) \(=(\mathrm{x}+7) \hat{\mathrm{i}}+(\mathrm{y}+3) \hat{\mathrm{j}}+(\mathrm{z}+12) \hat{\mathrm{k}}\) \(x+7=6\) i.e. \(x=-1, y+3=3\) i.e. \(y=0\) and \(z+12=15\) i.e. \(z=3\) So, \((x, y, z)=(-1,0,3)\)
MHT CET-2010
Vector Algebra
87678
If \(2 \vec{a}+3 \vec{b}-5 \vec{c}=0\), then the ratio in which \(C\) divided \(A B\) is
1 \(2: 3\) internally
2 \(2: 3\) externally
3 \(3: 2\) internally
4 \(3: 2\) externally
Explanation:
(C) : Given, \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-5 \overrightarrow{\mathrm{c}}=0\) \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}=5 \overrightarrow{\mathrm{c}}\) \(\frac{2 \vec{a}+3 \vec{b}}{5}=\vec{c}, \quad \frac{2 \vec{a}+3 \vec{b}}{3+2}=\vec{c}\) \(\frac{2\left(\overrightarrow{\mathrm{a}}+\frac{3}{2} \overrightarrow{\mathrm{b}}\right)}{2\left(\frac{3}{2}+1\right)}=\overrightarrow{\mathrm{c}} \text {, }\) \(\frac{\left(\vec{a}+\frac{3}{2} \vec{b}\right)}{\left(1+\frac{3}{2}\right)}=\vec{c}\) We know that, \(c=\frac{a+\lambda b}{1+\lambda}\) From comparing equation (i) \& (ii), we get - \(\lambda=\frac{3}{2}\) Hence, the point \(C\) divides \(A B\) internally in the ratio of \(3: 2\).
MHT CET-2009
Vector Algebra
87681
If the position vectors of the vertices \(A, B, C\) of a triangle \(\mathrm{ABC}\) are \(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}+6 \hat{\mathbf{j}}+6 \hat{\mathrm{k}}\) and \(-4 \hat{i}+9 \hat{j}+6 \hat{k}\) respectively, the triangle is :
1 equilateral
2 isosceles
3 scalene
4 right angled and isosceles also
Explanation:
(D) : Given, The position vectors of the vertices, \(\overrightarrow{\mathrm{OA}}=7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}},\) \(\overrightarrow{\mathrm{OC}}=-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\) \(\overrightarrow{\mathrm{AB}}=(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\sqrt{1+1+16}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}\) \(\overrightarrow{\mathrm{BC}}=(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{BC}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(\mid \overrightarrow{\mathrm{BC}}=\sqrt{(-3)^2+(3)^2}\) \(\overrightarrow{\mathrm{BC}}=\sqrt{18}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}\) \(\overrightarrow{\mathrm{CA}}=(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})-(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{CA}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{(4)^2+(-2)^2+(4)^2}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{36}=6\) \(3 \sqrt{2}, 3 \sqrt{2}\) \& 6 are sides of a right angled \(\Delta\). \(|\mathrm{AB}|^2+|\mathrm{BC}|^2=|\mathrm{AC}|^2\) \(\because(3 \sqrt{2})^2+(3 \sqrt{2})^2=36\) Hence, the \(\triangle A B C\) is a right-angled and isosceles also.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Vector Algebra
87676
If the position vectors of the vertices \(A, B\) and \(C\) are \(6 \hat{i}, 6 \hat{j}\) and \(\hat{k}\) respectively with respect to origin \(O\), the volume of the tetrahedron \(O A B C\) is
1 6
2 3
3 \(\frac{1}{6}\)
4 \(\frac{1}{3}\)
Explanation:
(A) : Given, \(\overrightarrow{\mathrm{OA}}=6 \hat{\mathrm{i}}=6 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OB}}=6 \hat{\mathrm{j}}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{k}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) The position vectors of \(A, B, C\), with respect to origin respectively. \(\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\left|\begin{array}{lll} 6 &0& 0\\ 0& 6& 0\\ 0& 0& 1 \end{array}\right|=6(6-0)-0+0=36\) Now, volume of tetrahedron \(=\frac{1}{6}\left[\begin{array}{lll} \overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}} \end{array}\right]=\frac{1}{6}(36)=6\)
MHT CET-2012
Vector Algebra
87677
If \(A(x, 2,8), B(3, y, 4)\) and \(C(4,1, z)\) are vertices of \(\triangle \mathrm{ABC}\) and \(\mathrm{G}(2,1,5)\) is the centroid then the values of \(x, y\) and \(z\) are respectively
1 \((1,0,2)\)
2 \((-1,0,2)\)
3 \((1,0,3)\)
4 \((-1,0,3)\)
Explanation:
(D) : Let, \(\vec{a}, \vec{b}, \vec{c}, \vec{g}\) be the position vectors of \(A, B, C\) and \(G\) respectively. \(\therefore \quad \vec{a}=x \hat{i}+2 \hat{j}+8 \hat{k}\) \(\vec{b}=3 \hat{i}+y \hat{j}+4 \hat{k}\) \(\vec{c}=4 \hat{i}+\hat{j}+z \hat{k}\) \(\vec{g}=2 \hat{i}+\hat{j}+5 \hat{k}\) \(A s G\) is the centroid of \(\triangle A B C, \vec{g}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\) i.e. \(3 \overrightarrow{\mathrm{g}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\) \(3(2 \hat{i}+\hat{j}+5 \hat{k})=(x \hat{i}+2 \hat{j}+8 \hat{k})+(3 \hat{i}+y \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}+z \hat{k})\) \(6 \hat{i}+3 \hat{j}+15 \hat{k}=(x+3+4) \hat{i}+(2+y+1) \hat{j}+(8+4+z) \hat{k}\) \(=(\mathrm{x}+7) \hat{\mathrm{i}}+(\mathrm{y}+3) \hat{\mathrm{j}}+(\mathrm{z}+12) \hat{\mathrm{k}}\) \(x+7=6\) i.e. \(x=-1, y+3=3\) i.e. \(y=0\) and \(z+12=15\) i.e. \(z=3\) So, \((x, y, z)=(-1,0,3)\)
MHT CET-2010
Vector Algebra
87678
If \(2 \vec{a}+3 \vec{b}-5 \vec{c}=0\), then the ratio in which \(C\) divided \(A B\) is
1 \(2: 3\) internally
2 \(2: 3\) externally
3 \(3: 2\) internally
4 \(3: 2\) externally
Explanation:
(C) : Given, \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-5 \overrightarrow{\mathrm{c}}=0\) \(2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}=5 \overrightarrow{\mathrm{c}}\) \(\frac{2 \vec{a}+3 \vec{b}}{5}=\vec{c}, \quad \frac{2 \vec{a}+3 \vec{b}}{3+2}=\vec{c}\) \(\frac{2\left(\overrightarrow{\mathrm{a}}+\frac{3}{2} \overrightarrow{\mathrm{b}}\right)}{2\left(\frac{3}{2}+1\right)}=\overrightarrow{\mathrm{c}} \text {, }\) \(\frac{\left(\vec{a}+\frac{3}{2} \vec{b}\right)}{\left(1+\frac{3}{2}\right)}=\vec{c}\) We know that, \(c=\frac{a+\lambda b}{1+\lambda}\) From comparing equation (i) \& (ii), we get - \(\lambda=\frac{3}{2}\) Hence, the point \(C\) divides \(A B\) internally in the ratio of \(3: 2\).
MHT CET-2009
Vector Algebra
87681
If the position vectors of the vertices \(A, B, C\) of a triangle \(\mathrm{ABC}\) are \(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}+6 \hat{\mathbf{j}}+6 \hat{\mathrm{k}}\) and \(-4 \hat{i}+9 \hat{j}+6 \hat{k}\) respectively, the triangle is :
1 equilateral
2 isosceles
3 scalene
4 right angled and isosceles also
Explanation:
(D) : Given, The position vectors of the vertices, \(\overrightarrow{\mathrm{OA}}=7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}},\) \(\overrightarrow{\mathrm{OC}}=-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\) \(\overrightarrow{\mathrm{AB}}=(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AB}}=\sqrt{1+1+16}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}\) \(\overrightarrow{\mathrm{BC}}=(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})-(-\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{BC}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) \(\mid \overrightarrow{\mathrm{BC}}=\sqrt{(-3)^2+(3)^2}\) \(\overrightarrow{\mathrm{BC}}=\sqrt{18}=3 \sqrt{2}\) \(\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}\) \(\overrightarrow{\mathrm{CA}}=(7 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})-(-4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\) \(\overrightarrow{\mathrm{CA}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{(4)^2+(-2)^2+(4)^2}\) \(|\overrightarrow{\mathrm{CA}}|=\sqrt{36}=6\) \(3 \sqrt{2}, 3 \sqrt{2}\) \& 6 are sides of a right angled \(\Delta\). \(|\mathrm{AB}|^2+|\mathrm{BC}|^2=|\mathrm{AC}|^2\) \(\because(3 \sqrt{2})^2+(3 \sqrt{2})^2=36\) Hence, the \(\triangle A B C\) is a right-angled and isosceles also.
