(A) :Given, Let unit vector be \(\hat{a}=x \hat{i}+y \hat{j}+z \hat{k}\), Which is perpendicular to \(-\hat{i}+2 \hat{j}+2 \hat{k}\). So, \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})=0\) \(-x \hat{i} \cdot \hat{i}+2 y \hat{j} \cdot \hat{j}+2 z \hat{k} \cdot \hat{k}=0\) \(\because(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\) \(\quad-x+2 y+2 z=0\) Since, unit vector ( \(x \hat{i}+y \hat{j}+z \hat{k})\) makes equal angles with \(x\) and \(y\) axes :- \(\therefore \cos \theta=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(x \hat{i})}{\sqrt{x^2+y^2+z^2} \sqrt{x^2}}=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(y \hat{j})}{\sqrt{x^2+y^2+z^2} \sqrt{y^2}}\) \(\frac{x^2}{x}=\frac{y^2}{y}\) \(x=y\) \(\therefore\) From (i), equation:- \(-x+2 x+2 z =0\) \(2 z =-x\) \(z =-\frac{x}{2}\) Now, \(x^2+y^2+z^2 =1\) \(x^2+x^2+\frac{x^2}{4} =1\) \(x =\frac{2}{3}\) Also, \(\vec{a}=x \hat{i}+x \hat{j}-\frac{x}{2} \hat{k}=\frac{x}{2}[2 \hat{i}+2 \hat{j}-\hat{k}]\) \(=\frac{2}{2 \cdot 3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}] \quad\left(\because \mathrm{x}=\frac{2}{3}\right)\) \(=\frac{1}{3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}]\)
SRM JEE-2015
Vector Algebra
87982
If \(\theta\) be the angle between vectors \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+\hat{k}\), then \(\cos \theta\) equals
87997
If \(|\vec{a} \times \vec{b}|=5\) and \(|\vec{a} \cdot \vec{b}|=3\), then \(|\vec{a}|^2|\vec{b}|^2\) is equal to
1 16
2 31
3 25
4 34
Explanation:
(D) : Given, \(|\vec{a} \times \vec{b}|=5 \text { and }|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a} \times \vec{b}|=5\) \(|\vec{a}||\vec{b}| \sin \theta=5\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta=25\) \(|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a}||\vec{b}| \cos \theta=3\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=9\) On adding equation (i) and (ii) \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta+|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=25+9\) \(|\vec{a}|^2|\vec{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=34\) \(|\vec{a}|^2|\vec{b}|^2=34 \quad\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
COMEDK-2012
Vector Algebra
87983
If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\vec{a}+2 \vec{b}+3 \vec{c}\), \(\lambda \vec{b}+4 \vec{c}\) and \((2 \lambda-1) \vec{c}\) are non coplanar for
1 no value of \(\lambda\)
2 all except one value of \(\lambda\)
3 all except two values of \(\lambda\)
4 all values of \(\lambda\)
Explanation:
(C): Given that, Vectors \(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}, \quad \lambda \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{c}}\) and \((2 \lambda-1) \overrightarrow{\mathrm{c}} \quad\) are coplanar if \(\left|\begin{array}{ccc}1 &2 &3 \\ 0& \lambda& 4 \\ 0& 0 &2 \lambda-1\end{array}\right|=0\) \(\lambda(2 \lambda-1)=0\) \(\lambda=0 \text { or } 2 \lambda-1=0\) \(\quad \lambda=\frac{1}{2}\) \(\text { or } \frac{1}{2} \text { Forces are non coplanar for all } \lambda \text {, excepts }\) \(\lambda=0, \frac{1}{2}\) or \(\frac{1}{2}\) Forces are non coplanar for all \(\lambda\), excepts \(\lambda=0, \frac{1}{2}\)
(A) :Given, Let unit vector be \(\hat{a}=x \hat{i}+y \hat{j}+z \hat{k}\), Which is perpendicular to \(-\hat{i}+2 \hat{j}+2 \hat{k}\). So, \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})=0\) \(-x \hat{i} \cdot \hat{i}+2 y \hat{j} \cdot \hat{j}+2 z \hat{k} \cdot \hat{k}=0\) \(\because(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\) \(\quad-x+2 y+2 z=0\) Since, unit vector ( \(x \hat{i}+y \hat{j}+z \hat{k})\) makes equal angles with \(x\) and \(y\) axes :- \(\therefore \cos \theta=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(x \hat{i})}{\sqrt{x^2+y^2+z^2} \sqrt{x^2}}=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(y \hat{j})}{\sqrt{x^2+y^2+z^2} \sqrt{y^2}}\) \(\frac{x^2}{x}=\frac{y^2}{y}\) \(x=y\) \(\therefore\) From (i), equation:- \(-x+2 x+2 z =0\) \(2 z =-x\) \(z =-\frac{x}{2}\) Now, \(x^2+y^2+z^2 =1\) \(x^2+x^2+\frac{x^2}{4} =1\) \(x =\frac{2}{3}\) Also, \(\vec{a}=x \hat{i}+x \hat{j}-\frac{x}{2} \hat{k}=\frac{x}{2}[2 \hat{i}+2 \hat{j}-\hat{k}]\) \(=\frac{2}{2 \cdot 3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}] \quad\left(\because \mathrm{x}=\frac{2}{3}\right)\) \(=\frac{1}{3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}]\)
SRM JEE-2015
Vector Algebra
87982
If \(\theta\) be the angle between vectors \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+\hat{k}\), then \(\cos \theta\) equals
87997
If \(|\vec{a} \times \vec{b}|=5\) and \(|\vec{a} \cdot \vec{b}|=3\), then \(|\vec{a}|^2|\vec{b}|^2\) is equal to
1 16
2 31
3 25
4 34
Explanation:
(D) : Given, \(|\vec{a} \times \vec{b}|=5 \text { and }|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a} \times \vec{b}|=5\) \(|\vec{a}||\vec{b}| \sin \theta=5\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta=25\) \(|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a}||\vec{b}| \cos \theta=3\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=9\) On adding equation (i) and (ii) \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta+|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=25+9\) \(|\vec{a}|^2|\vec{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=34\) \(|\vec{a}|^2|\vec{b}|^2=34 \quad\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
COMEDK-2012
Vector Algebra
87983
If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\vec{a}+2 \vec{b}+3 \vec{c}\), \(\lambda \vec{b}+4 \vec{c}\) and \((2 \lambda-1) \vec{c}\) are non coplanar for
1 no value of \(\lambda\)
2 all except one value of \(\lambda\)
3 all except two values of \(\lambda\)
4 all values of \(\lambda\)
Explanation:
(C): Given that, Vectors \(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}, \quad \lambda \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{c}}\) and \((2 \lambda-1) \overrightarrow{\mathrm{c}} \quad\) are coplanar if \(\left|\begin{array}{ccc}1 &2 &3 \\ 0& \lambda& 4 \\ 0& 0 &2 \lambda-1\end{array}\right|=0\) \(\lambda(2 \lambda-1)=0\) \(\lambda=0 \text { or } 2 \lambda-1=0\) \(\quad \lambda=\frac{1}{2}\) \(\text { or } \frac{1}{2} \text { Forces are non coplanar for all } \lambda \text {, excepts }\) \(\lambda=0, \frac{1}{2}\) or \(\frac{1}{2}\) Forces are non coplanar for all \(\lambda\), excepts \(\lambda=0, \frac{1}{2}\)
(A) :Given, Let unit vector be \(\hat{a}=x \hat{i}+y \hat{j}+z \hat{k}\), Which is perpendicular to \(-\hat{i}+2 \hat{j}+2 \hat{k}\). So, \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})=0\) \(-x \hat{i} \cdot \hat{i}+2 y \hat{j} \cdot \hat{j}+2 z \hat{k} \cdot \hat{k}=0\) \(\because(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\) \(\quad-x+2 y+2 z=0\) Since, unit vector ( \(x \hat{i}+y \hat{j}+z \hat{k})\) makes equal angles with \(x\) and \(y\) axes :- \(\therefore \cos \theta=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(x \hat{i})}{\sqrt{x^2+y^2+z^2} \sqrt{x^2}}=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(y \hat{j})}{\sqrt{x^2+y^2+z^2} \sqrt{y^2}}\) \(\frac{x^2}{x}=\frac{y^2}{y}\) \(x=y\) \(\therefore\) From (i), equation:- \(-x+2 x+2 z =0\) \(2 z =-x\) \(z =-\frac{x}{2}\) Now, \(x^2+y^2+z^2 =1\) \(x^2+x^2+\frac{x^2}{4} =1\) \(x =\frac{2}{3}\) Also, \(\vec{a}=x \hat{i}+x \hat{j}-\frac{x}{2} \hat{k}=\frac{x}{2}[2 \hat{i}+2 \hat{j}-\hat{k}]\) \(=\frac{2}{2 \cdot 3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}] \quad\left(\because \mathrm{x}=\frac{2}{3}\right)\) \(=\frac{1}{3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}]\)
SRM JEE-2015
Vector Algebra
87982
If \(\theta\) be the angle between vectors \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+\hat{k}\), then \(\cos \theta\) equals
87997
If \(|\vec{a} \times \vec{b}|=5\) and \(|\vec{a} \cdot \vec{b}|=3\), then \(|\vec{a}|^2|\vec{b}|^2\) is equal to
1 16
2 31
3 25
4 34
Explanation:
(D) : Given, \(|\vec{a} \times \vec{b}|=5 \text { and }|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a} \times \vec{b}|=5\) \(|\vec{a}||\vec{b}| \sin \theta=5\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta=25\) \(|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a}||\vec{b}| \cos \theta=3\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=9\) On adding equation (i) and (ii) \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta+|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=25+9\) \(|\vec{a}|^2|\vec{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=34\) \(|\vec{a}|^2|\vec{b}|^2=34 \quad\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
COMEDK-2012
Vector Algebra
87983
If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\vec{a}+2 \vec{b}+3 \vec{c}\), \(\lambda \vec{b}+4 \vec{c}\) and \((2 \lambda-1) \vec{c}\) are non coplanar for
1 no value of \(\lambda\)
2 all except one value of \(\lambda\)
3 all except two values of \(\lambda\)
4 all values of \(\lambda\)
Explanation:
(C): Given that, Vectors \(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}, \quad \lambda \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{c}}\) and \((2 \lambda-1) \overrightarrow{\mathrm{c}} \quad\) are coplanar if \(\left|\begin{array}{ccc}1 &2 &3 \\ 0& \lambda& 4 \\ 0& 0 &2 \lambda-1\end{array}\right|=0\) \(\lambda(2 \lambda-1)=0\) \(\lambda=0 \text { or } 2 \lambda-1=0\) \(\quad \lambda=\frac{1}{2}\) \(\text { or } \frac{1}{2} \text { Forces are non coplanar for all } \lambda \text {, excepts }\) \(\lambda=0, \frac{1}{2}\) or \(\frac{1}{2}\) Forces are non coplanar for all \(\lambda\), excepts \(\lambda=0, \frac{1}{2}\)
(A) :Given, Let unit vector be \(\hat{a}=x \hat{i}+y \hat{j}+z \hat{k}\), Which is perpendicular to \(-\hat{i}+2 \hat{j}+2 \hat{k}\). So, \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})=0\) \(-x \hat{i} \cdot \hat{i}+2 y \hat{j} \cdot \hat{j}+2 z \hat{k} \cdot \hat{k}=0\) \(\because(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\) \(\quad-x+2 y+2 z=0\) Since, unit vector ( \(x \hat{i}+y \hat{j}+z \hat{k})\) makes equal angles with \(x\) and \(y\) axes :- \(\therefore \cos \theta=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(x \hat{i})}{\sqrt{x^2+y^2+z^2} \sqrt{x^2}}=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(y \hat{j})}{\sqrt{x^2+y^2+z^2} \sqrt{y^2}}\) \(\frac{x^2}{x}=\frac{y^2}{y}\) \(x=y\) \(\therefore\) From (i), equation:- \(-x+2 x+2 z =0\) \(2 z =-x\) \(z =-\frac{x}{2}\) Now, \(x^2+y^2+z^2 =1\) \(x^2+x^2+\frac{x^2}{4} =1\) \(x =\frac{2}{3}\) Also, \(\vec{a}=x \hat{i}+x \hat{j}-\frac{x}{2} \hat{k}=\frac{x}{2}[2 \hat{i}+2 \hat{j}-\hat{k}]\) \(=\frac{2}{2 \cdot 3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}] \quad\left(\because \mathrm{x}=\frac{2}{3}\right)\) \(=\frac{1}{3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}]\)
SRM JEE-2015
Vector Algebra
87982
If \(\theta\) be the angle between vectors \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+\hat{k}\), then \(\cos \theta\) equals
87997
If \(|\vec{a} \times \vec{b}|=5\) and \(|\vec{a} \cdot \vec{b}|=3\), then \(|\vec{a}|^2|\vec{b}|^2\) is equal to
1 16
2 31
3 25
4 34
Explanation:
(D) : Given, \(|\vec{a} \times \vec{b}|=5 \text { and }|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a} \times \vec{b}|=5\) \(|\vec{a}||\vec{b}| \sin \theta=5\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta=25\) \(|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a}||\vec{b}| \cos \theta=3\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=9\) On adding equation (i) and (ii) \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta+|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=25+9\) \(|\vec{a}|^2|\vec{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=34\) \(|\vec{a}|^2|\vec{b}|^2=34 \quad\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
COMEDK-2012
Vector Algebra
87983
If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\vec{a}+2 \vec{b}+3 \vec{c}\), \(\lambda \vec{b}+4 \vec{c}\) and \((2 \lambda-1) \vec{c}\) are non coplanar for
1 no value of \(\lambda\)
2 all except one value of \(\lambda\)
3 all except two values of \(\lambda\)
4 all values of \(\lambda\)
Explanation:
(C): Given that, Vectors \(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}, \quad \lambda \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{c}}\) and \((2 \lambda-1) \overrightarrow{\mathrm{c}} \quad\) are coplanar if \(\left|\begin{array}{ccc}1 &2 &3 \\ 0& \lambda& 4 \\ 0& 0 &2 \lambda-1\end{array}\right|=0\) \(\lambda(2 \lambda-1)=0\) \(\lambda=0 \text { or } 2 \lambda-1=0\) \(\quad \lambda=\frac{1}{2}\) \(\text { or } \frac{1}{2} \text { Forces are non coplanar for all } \lambda \text {, excepts }\) \(\lambda=0, \frac{1}{2}\) or \(\frac{1}{2}\) Forces are non coplanar for all \(\lambda\), excepts \(\lambda=0, \frac{1}{2}\)
(A) :Given, Let unit vector be \(\hat{a}=x \hat{i}+y \hat{j}+z \hat{k}\), Which is perpendicular to \(-\hat{i}+2 \hat{j}+2 \hat{k}\). So, \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})=0\) \(-x \hat{i} \cdot \hat{i}+2 y \hat{j} \cdot \hat{j}+2 z \hat{k} \cdot \hat{k}=0\) \(\because(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\) \(\quad-x+2 y+2 z=0\) Since, unit vector ( \(x \hat{i}+y \hat{j}+z \hat{k})\) makes equal angles with \(x\) and \(y\) axes :- \(\therefore \cos \theta=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(x \hat{i})}{\sqrt{x^2+y^2+z^2} \sqrt{x^2}}=\frac{(x \hat{i}+y \hat{j}+z \hat{k})(y \hat{j})}{\sqrt{x^2+y^2+z^2} \sqrt{y^2}}\) \(\frac{x^2}{x}=\frac{y^2}{y}\) \(x=y\) \(\therefore\) From (i), equation:- \(-x+2 x+2 z =0\) \(2 z =-x\) \(z =-\frac{x}{2}\) Now, \(x^2+y^2+z^2 =1\) \(x^2+x^2+\frac{x^2}{4} =1\) \(x =\frac{2}{3}\) Also, \(\vec{a}=x \hat{i}+x \hat{j}-\frac{x}{2} \hat{k}=\frac{x}{2}[2 \hat{i}+2 \hat{j}-\hat{k}]\) \(=\frac{2}{2 \cdot 3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}] \quad\left(\because \mathrm{x}=\frac{2}{3}\right)\) \(=\frac{1}{3}[2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}]\)
SRM JEE-2015
Vector Algebra
87982
If \(\theta\) be the angle between vectors \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+\hat{k}\), then \(\cos \theta\) equals
87997
If \(|\vec{a} \times \vec{b}|=5\) and \(|\vec{a} \cdot \vec{b}|=3\), then \(|\vec{a}|^2|\vec{b}|^2\) is equal to
1 16
2 31
3 25
4 34
Explanation:
(D) : Given, \(|\vec{a} \times \vec{b}|=5 \text { and }|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a} \times \vec{b}|=5\) \(|\vec{a}||\vec{b}| \sin \theta=5\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta=25\) \(|\vec{a} \cdot \vec{b}|=3\) \(|\vec{a}||\vec{b}| \cos \theta=3\) Squaring on both side \(|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=9\) On adding equation (i) and (ii) \(|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta+|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=25+9\) \(|\vec{a}|^2|\vec{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=34\) \(|\vec{a}|^2|\vec{b}|^2=34 \quad\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]\)
COMEDK-2012
Vector Algebra
87983
If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\vec{a}+2 \vec{b}+3 \vec{c}\), \(\lambda \vec{b}+4 \vec{c}\) and \((2 \lambda-1) \vec{c}\) are non coplanar for
1 no value of \(\lambda\)
2 all except one value of \(\lambda\)
3 all except two values of \(\lambda\)
4 all values of \(\lambda\)
Explanation:
(C): Given that, Vectors \(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}, \quad \lambda \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{c}}\) and \((2 \lambda-1) \overrightarrow{\mathrm{c}} \quad\) are coplanar if \(\left|\begin{array}{ccc}1 &2 &3 \\ 0& \lambda& 4 \\ 0& 0 &2 \lambda-1\end{array}\right|=0\) \(\lambda(2 \lambda-1)=0\) \(\lambda=0 \text { or } 2 \lambda-1=0\) \(\quad \lambda=\frac{1}{2}\) \(\text { or } \frac{1}{2} \text { Forces are non coplanar for all } \lambda \text {, excepts }\) \(\lambda=0, \frac{1}{2}\) or \(\frac{1}{2}\) Forces are non coplanar for all \(\lambda\), excepts \(\lambda=0, \frac{1}{2}\)