(B) : It is given that, \(\overrightarrow{\mathrm{u}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{v}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} &\hat{\mathrm{j}} &\hat{\mathrm{k}}\\ 2 &2 &-1\\ 6 &-3& 2 \end{array}\right|\) \(=\hat{\mathrm{i}}(4-3)-\hat{\mathrm{j}}(4+6)+\hat{\mathrm{k}}(-6-12)\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|=\sqrt{1+100+324}=\sqrt{425}=5 \sqrt{17}\) Required unit vector perpendicular to \(\overrightarrow{\mathrm{u}}\) and \(\overrightarrow{\mathrm{v}}\), \(\frac{(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})}{|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|} =\frac{\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}}{5 \sqrt{17}}\) \(=\frac{1}{\sqrt{17}}\left(\frac{1}{5} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\frac{18}{5} \hat{\mathrm{k}}\right)\)
CG PET-2006
Vector Algebra
88021
If \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) are vectors in space, then the value of \((2 a+b) \cdot[(a \times b) \times(a-\) 2b)] is
1 0
2 1
3 5
4 4
Explanation:
(C) : Given, \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) Now, \((2 a+b) \cdot[(a \times b) \times(a-2 b)]\) \(= (2 a+b) \cdot[(a \times b) \times a-2(a \times b) \times b]\) \(= (2 a+b) \cdot[(a \cdot a) b-(b \cdot a) a\) \(-2\{(a \cdot b) b-(b \cdot b) a\}]\) \(= (2 a+b)\left[|a|^2 b-0 \cdot a-2\left\{0 \cdot b-|b|^2 a\right\}\right]\) \(= (2 a+b) \cdot[1 \cdot b+2 \cdot 1 \cdot a] \quad[\because|a|=1,|b|=1]\) \(= (2 a+b)(2 a+b)=4|a|^2+|b|^2+4 a \cdot b\) \(= 4 \times 1+1+4 \times 0 \quad[\because|a|=1=|b| \text { and } a \cdot b=0]\) \(= 4+1=5\)
CG PET-2015
Vector Algebra
88023
A unit vector \(\vec{a}\) makes angles \(\pi / 4\) with \(\hat{\mathbf{i}}, \pi / 3\) with \(\hat{\mathbf{j}}\) and an acute angle \(\theta\) with \(\hat{\mathbf{k}}\), then \(\boldsymbol{\theta}\) and \(\vec{a}\) are
88024
If \(a=3 \hat{i}-4 \hat{j}+5 \hat{k}, b=\hat{i}+\hat{j}+\hat{k}\) and \(c=-2 \hat{i}+3 \hat{j}-5 \hat{k}\) and if [.] is the least integer function, then \([a+b+c]\) is equal to
(B) : It is given that, \(\overrightarrow{\mathrm{u}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{v}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} &\hat{\mathrm{j}} &\hat{\mathrm{k}}\\ 2 &2 &-1\\ 6 &-3& 2 \end{array}\right|\) \(=\hat{\mathrm{i}}(4-3)-\hat{\mathrm{j}}(4+6)+\hat{\mathrm{k}}(-6-12)\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|=\sqrt{1+100+324}=\sqrt{425}=5 \sqrt{17}\) Required unit vector perpendicular to \(\overrightarrow{\mathrm{u}}\) and \(\overrightarrow{\mathrm{v}}\), \(\frac{(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})}{|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|} =\frac{\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}}{5 \sqrt{17}}\) \(=\frac{1}{\sqrt{17}}\left(\frac{1}{5} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\frac{18}{5} \hat{\mathrm{k}}\right)\)
CG PET-2006
Vector Algebra
88021
If \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) are vectors in space, then the value of \((2 a+b) \cdot[(a \times b) \times(a-\) 2b)] is
1 0
2 1
3 5
4 4
Explanation:
(C) : Given, \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) Now, \((2 a+b) \cdot[(a \times b) \times(a-2 b)]\) \(= (2 a+b) \cdot[(a \times b) \times a-2(a \times b) \times b]\) \(= (2 a+b) \cdot[(a \cdot a) b-(b \cdot a) a\) \(-2\{(a \cdot b) b-(b \cdot b) a\}]\) \(= (2 a+b)\left[|a|^2 b-0 \cdot a-2\left\{0 \cdot b-|b|^2 a\right\}\right]\) \(= (2 a+b) \cdot[1 \cdot b+2 \cdot 1 \cdot a] \quad[\because|a|=1,|b|=1]\) \(= (2 a+b)(2 a+b)=4|a|^2+|b|^2+4 a \cdot b\) \(= 4 \times 1+1+4 \times 0 \quad[\because|a|=1=|b| \text { and } a \cdot b=0]\) \(= 4+1=5\)
CG PET-2015
Vector Algebra
88023
A unit vector \(\vec{a}\) makes angles \(\pi / 4\) with \(\hat{\mathbf{i}}, \pi / 3\) with \(\hat{\mathbf{j}}\) and an acute angle \(\theta\) with \(\hat{\mathbf{k}}\), then \(\boldsymbol{\theta}\) and \(\vec{a}\) are
88024
If \(a=3 \hat{i}-4 \hat{j}+5 \hat{k}, b=\hat{i}+\hat{j}+\hat{k}\) and \(c=-2 \hat{i}+3 \hat{j}-5 \hat{k}\) and if [.] is the least integer function, then \([a+b+c]\) is equal to
(B) : It is given that, \(\overrightarrow{\mathrm{u}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{v}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} &\hat{\mathrm{j}} &\hat{\mathrm{k}}\\ 2 &2 &-1\\ 6 &-3& 2 \end{array}\right|\) \(=\hat{\mathrm{i}}(4-3)-\hat{\mathrm{j}}(4+6)+\hat{\mathrm{k}}(-6-12)\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|=\sqrt{1+100+324}=\sqrt{425}=5 \sqrt{17}\) Required unit vector perpendicular to \(\overrightarrow{\mathrm{u}}\) and \(\overrightarrow{\mathrm{v}}\), \(\frac{(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})}{|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|} =\frac{\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}}{5 \sqrt{17}}\) \(=\frac{1}{\sqrt{17}}\left(\frac{1}{5} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\frac{18}{5} \hat{\mathrm{k}}\right)\)
CG PET-2006
Vector Algebra
88021
If \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) are vectors in space, then the value of \((2 a+b) \cdot[(a \times b) \times(a-\) 2b)] is
1 0
2 1
3 5
4 4
Explanation:
(C) : Given, \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) Now, \((2 a+b) \cdot[(a \times b) \times(a-2 b)]\) \(= (2 a+b) \cdot[(a \times b) \times a-2(a \times b) \times b]\) \(= (2 a+b) \cdot[(a \cdot a) b-(b \cdot a) a\) \(-2\{(a \cdot b) b-(b \cdot b) a\}]\) \(= (2 a+b)\left[|a|^2 b-0 \cdot a-2\left\{0 \cdot b-|b|^2 a\right\}\right]\) \(= (2 a+b) \cdot[1 \cdot b+2 \cdot 1 \cdot a] \quad[\because|a|=1,|b|=1]\) \(= (2 a+b)(2 a+b)=4|a|^2+|b|^2+4 a \cdot b\) \(= 4 \times 1+1+4 \times 0 \quad[\because|a|=1=|b| \text { and } a \cdot b=0]\) \(= 4+1=5\)
CG PET-2015
Vector Algebra
88023
A unit vector \(\vec{a}\) makes angles \(\pi / 4\) with \(\hat{\mathbf{i}}, \pi / 3\) with \(\hat{\mathbf{j}}\) and an acute angle \(\theta\) with \(\hat{\mathbf{k}}\), then \(\boldsymbol{\theta}\) and \(\vec{a}\) are
88024
If \(a=3 \hat{i}-4 \hat{j}+5 \hat{k}, b=\hat{i}+\hat{j}+\hat{k}\) and \(c=-2 \hat{i}+3 \hat{j}-5 \hat{k}\) and if [.] is the least integer function, then \([a+b+c]\) is equal to
(B) : It is given that, \(\overrightarrow{\mathrm{u}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{v}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} &\hat{\mathrm{j}} &\hat{\mathrm{k}}\\ 2 &2 &-1\\ 6 &-3& 2 \end{array}\right|\) \(=\hat{\mathrm{i}}(4-3)-\hat{\mathrm{j}}(4+6)+\hat{\mathrm{k}}(-6-12)\) \(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}\) \(|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|=\sqrt{1+100+324}=\sqrt{425}=5 \sqrt{17}\) Required unit vector perpendicular to \(\overrightarrow{\mathrm{u}}\) and \(\overrightarrow{\mathrm{v}}\), \(\frac{(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}})}{|\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|} =\frac{\hat{\mathrm{i}}-10 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}}{5 \sqrt{17}}\) \(=\frac{1}{\sqrt{17}}\left(\frac{1}{5} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\frac{18}{5} \hat{\mathrm{k}}\right)\)
CG PET-2006
Vector Algebra
88021
If \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) are vectors in space, then the value of \((2 a+b) \cdot[(a \times b) \times(a-\) 2b)] is
1 0
2 1
3 5
4 4
Explanation:
(C) : Given, \(a=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\) and \(b=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}\) Now, \((2 a+b) \cdot[(a \times b) \times(a-2 b)]\) \(= (2 a+b) \cdot[(a \times b) \times a-2(a \times b) \times b]\) \(= (2 a+b) \cdot[(a \cdot a) b-(b \cdot a) a\) \(-2\{(a \cdot b) b-(b \cdot b) a\}]\) \(= (2 a+b)\left[|a|^2 b-0 \cdot a-2\left\{0 \cdot b-|b|^2 a\right\}\right]\) \(= (2 a+b) \cdot[1 \cdot b+2 \cdot 1 \cdot a] \quad[\because|a|=1,|b|=1]\) \(= (2 a+b)(2 a+b)=4|a|^2+|b|^2+4 a \cdot b\) \(= 4 \times 1+1+4 \times 0 \quad[\because|a|=1=|b| \text { and } a \cdot b=0]\) \(= 4+1=5\)
CG PET-2015
Vector Algebra
88023
A unit vector \(\vec{a}\) makes angles \(\pi / 4\) with \(\hat{\mathbf{i}}, \pi / 3\) with \(\hat{\mathbf{j}}\) and an acute angle \(\theta\) with \(\hat{\mathbf{k}}\), then \(\boldsymbol{\theta}\) and \(\vec{a}\) are
88024
If \(a=3 \hat{i}-4 \hat{j}+5 \hat{k}, b=\hat{i}+\hat{j}+\hat{k}\) and \(c=-2 \hat{i}+3 \hat{j}-5 \hat{k}\) and if [.] is the least integer function, then \([a+b+c]\) is equal to