87984
A vector of magnitude 5 and perpendicular to \((\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})\) is :
(A) : Given, \(\vec{A}=(\hat{i}-2 \hat{j}+\hat{k})\) and \(\vec{B}=(2 \hat{i}+\hat{j}-3 \hat{k})\) Now \(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} &\hat{\mathbf{j}}& \hat{\mathrm{k}} \\ 1 &-2 &1 \\ 2 &1& -3\end{array}\right|\) \(=\hat{i}(6-1)-\hat{j}(-3-2)+\hat{k}(1+4)=5 \hat{i}+5 \hat{j}+5 \hat{k}\) and \(|\vec{A} \times \vec{B}|=\sqrt{25+25+25}=5 \sqrt{3}\) We know that, \(\therefore\) unit vector along \(\vec{A} \times \vec{B}=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\) \(\frac{5(\hat{i}+\hat{j}+\hat{k})}{5 \sqrt{3}}=\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)\) \(\because\) vector of magnitude is 5 Thus, required vector \(=\frac{5 \sqrt{3}}{3}(\hat{i}+\hat{j}+\hat{k})\)
BITSAT-2006
Vector Algebra
87985
If vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar, then the value of \(a\) is
1 2
2 -2
3 -1
4 -4
Explanation:
(D) : Given vector \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\) \(\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{c}=3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar We know that \({[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=0}\) \(\left|\begin{array}{ccc} 2 &-1& 1\\ 1& 2& -3\\ 3 &a& 5 \end{array}\right|=0\) \(2(10+3 a)+1(5+9)+1(a-6)=0\) \(20+6 \mathrm{a}+14+\mathrm{a}-6=0\) \(7 \mathrm{a}+28=0\) \(\mathrm{a}=-4\)
BITSAT-2013
Vector Algebra
87986
The unit vector perpendicular to the vectors \(6 \hat{i}+2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-6 \hat{j}-2 \hat{k}\) is-
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Vector Algebra
87984
A vector of magnitude 5 and perpendicular to \((\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})\) is :
(A) : Given, \(\vec{A}=(\hat{i}-2 \hat{j}+\hat{k})\) and \(\vec{B}=(2 \hat{i}+\hat{j}-3 \hat{k})\) Now \(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} &\hat{\mathbf{j}}& \hat{\mathrm{k}} \\ 1 &-2 &1 \\ 2 &1& -3\end{array}\right|\) \(=\hat{i}(6-1)-\hat{j}(-3-2)+\hat{k}(1+4)=5 \hat{i}+5 \hat{j}+5 \hat{k}\) and \(|\vec{A} \times \vec{B}|=\sqrt{25+25+25}=5 \sqrt{3}\) We know that, \(\therefore\) unit vector along \(\vec{A} \times \vec{B}=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\) \(\frac{5(\hat{i}+\hat{j}+\hat{k})}{5 \sqrt{3}}=\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)\) \(\because\) vector of magnitude is 5 Thus, required vector \(=\frac{5 \sqrt{3}}{3}(\hat{i}+\hat{j}+\hat{k})\)
BITSAT-2006
Vector Algebra
87985
If vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar, then the value of \(a\) is
1 2
2 -2
3 -1
4 -4
Explanation:
(D) : Given vector \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\) \(\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{c}=3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar We know that \({[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=0}\) \(\left|\begin{array}{ccc} 2 &-1& 1\\ 1& 2& -3\\ 3 &a& 5 \end{array}\right|=0\) \(2(10+3 a)+1(5+9)+1(a-6)=0\) \(20+6 \mathrm{a}+14+\mathrm{a}-6=0\) \(7 \mathrm{a}+28=0\) \(\mathrm{a}=-4\)
BITSAT-2013
Vector Algebra
87986
The unit vector perpendicular to the vectors \(6 \hat{i}+2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-6 \hat{j}-2 \hat{k}\) is-
87984
A vector of magnitude 5 and perpendicular to \((\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})\) is :
(A) : Given, \(\vec{A}=(\hat{i}-2 \hat{j}+\hat{k})\) and \(\vec{B}=(2 \hat{i}+\hat{j}-3 \hat{k})\) Now \(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} &\hat{\mathbf{j}}& \hat{\mathrm{k}} \\ 1 &-2 &1 \\ 2 &1& -3\end{array}\right|\) \(=\hat{i}(6-1)-\hat{j}(-3-2)+\hat{k}(1+4)=5 \hat{i}+5 \hat{j}+5 \hat{k}\) and \(|\vec{A} \times \vec{B}|=\sqrt{25+25+25}=5 \sqrt{3}\) We know that, \(\therefore\) unit vector along \(\vec{A} \times \vec{B}=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\) \(\frac{5(\hat{i}+\hat{j}+\hat{k})}{5 \sqrt{3}}=\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)\) \(\because\) vector of magnitude is 5 Thus, required vector \(=\frac{5 \sqrt{3}}{3}(\hat{i}+\hat{j}+\hat{k})\)
BITSAT-2006
Vector Algebra
87985
If vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar, then the value of \(a\) is
1 2
2 -2
3 -1
4 -4
Explanation:
(D) : Given vector \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\) \(\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{c}=3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar We know that \({[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=0}\) \(\left|\begin{array}{ccc} 2 &-1& 1\\ 1& 2& -3\\ 3 &a& 5 \end{array}\right|=0\) \(2(10+3 a)+1(5+9)+1(a-6)=0\) \(20+6 \mathrm{a}+14+\mathrm{a}-6=0\) \(7 \mathrm{a}+28=0\) \(\mathrm{a}=-4\)
BITSAT-2013
Vector Algebra
87986
The unit vector perpendicular to the vectors \(6 \hat{i}+2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-6 \hat{j}-2 \hat{k}\) is-
87984
A vector of magnitude 5 and perpendicular to \((\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})\) is :
(A) : Given, \(\vec{A}=(\hat{i}-2 \hat{j}+\hat{k})\) and \(\vec{B}=(2 \hat{i}+\hat{j}-3 \hat{k})\) Now \(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} &\hat{\mathbf{j}}& \hat{\mathrm{k}} \\ 1 &-2 &1 \\ 2 &1& -3\end{array}\right|\) \(=\hat{i}(6-1)-\hat{j}(-3-2)+\hat{k}(1+4)=5 \hat{i}+5 \hat{j}+5 \hat{k}\) and \(|\vec{A} \times \vec{B}|=\sqrt{25+25+25}=5 \sqrt{3}\) We know that, \(\therefore\) unit vector along \(\vec{A} \times \vec{B}=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\) \(\frac{5(\hat{i}+\hat{j}+\hat{k})}{5 \sqrt{3}}=\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)\) \(\because\) vector of magnitude is 5 Thus, required vector \(=\frac{5 \sqrt{3}}{3}(\hat{i}+\hat{j}+\hat{k})\)
BITSAT-2006
Vector Algebra
87985
If vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar, then the value of \(a\) is
1 2
2 -2
3 -1
4 -4
Explanation:
(D) : Given vector \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\) \(\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}\) \(\vec{c}=3 \hat{i}+a \hat{j}+5 \hat{k}\) are coplanar We know that \({[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=0}\) \(\left|\begin{array}{ccc} 2 &-1& 1\\ 1& 2& -3\\ 3 &a& 5 \end{array}\right|=0\) \(2(10+3 a)+1(5+9)+1(a-6)=0\) \(20+6 \mathrm{a}+14+\mathrm{a}-6=0\) \(7 \mathrm{a}+28=0\) \(\mathrm{a}=-4\)
BITSAT-2013
Vector Algebra
87986
The unit vector perpendicular to the vectors \(6 \hat{i}+2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-6 \hat{j}-2 \hat{k}\) is-