87736
If three vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}-\mathbf{3} \hat{\mathbf{k}}\) and \(3 \hat{i}+\lambda \hat{i}+5 \hat{k}\) are coplanar, then the value of \(\lambda\) is
87737
The two vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+3 \hat{j}+5 \hat{k}\) represents the two sides \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) respectively of a \(\triangle \mathrm{ABC}\). The length of the median though \(A\) is
1 14
2 7
3 \(\sqrt{14}\)
4 \(\frac{\sqrt{14}}{2}\)
Explanation:
(C) : Given, \(\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) We know that, \(\text { Median } =\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+3 \hat{j}+5 \hat{k})}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{2 \hat{i}+4 \hat{j}+6 \hat{k}}{2}\) \(\overrightarrow{\mathrm{AD}} =\hat{i}+2 \hat{j}+3 \hat{k}\) \(\text { Length of Median } =\sqrt{1^2+2^2+3^2}\) \(=\sqrt{1+4+9}=\sqrt{14}\)
Karnataka CET-2020
Vector Algebra
87738
If \(\vec{a}=3, \vec{b}=4, \vec{c}=5\) each one of \(\vec{a}, \vec{b} \& \vec{c}\) is perpendicular to the sum of the remaining then \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}]\) is equal to
87736
If three vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}-\mathbf{3} \hat{\mathbf{k}}\) and \(3 \hat{i}+\lambda \hat{i}+5 \hat{k}\) are coplanar, then the value of \(\lambda\) is
87737
The two vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+3 \hat{j}+5 \hat{k}\) represents the two sides \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) respectively of a \(\triangle \mathrm{ABC}\). The length of the median though \(A\) is
1 14
2 7
3 \(\sqrt{14}\)
4 \(\frac{\sqrt{14}}{2}\)
Explanation:
(C) : Given, \(\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) We know that, \(\text { Median } =\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+3 \hat{j}+5 \hat{k})}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{2 \hat{i}+4 \hat{j}+6 \hat{k}}{2}\) \(\overrightarrow{\mathrm{AD}} =\hat{i}+2 \hat{j}+3 \hat{k}\) \(\text { Length of Median } =\sqrt{1^2+2^2+3^2}\) \(=\sqrt{1+4+9}=\sqrt{14}\)
Karnataka CET-2020
Vector Algebra
87738
If \(\vec{a}=3, \vec{b}=4, \vec{c}=5\) each one of \(\vec{a}, \vec{b} \& \vec{c}\) is perpendicular to the sum of the remaining then \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}]\) is equal to
87736
If three vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}-\mathbf{3} \hat{\mathbf{k}}\) and \(3 \hat{i}+\lambda \hat{i}+5 \hat{k}\) are coplanar, then the value of \(\lambda\) is
87737
The two vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+3 \hat{j}+5 \hat{k}\) represents the two sides \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) respectively of a \(\triangle \mathrm{ABC}\). The length of the median though \(A\) is
1 14
2 7
3 \(\sqrt{14}\)
4 \(\frac{\sqrt{14}}{2}\)
Explanation:
(C) : Given, \(\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) We know that, \(\text { Median } =\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+3 \hat{j}+5 \hat{k})}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{2 \hat{i}+4 \hat{j}+6 \hat{k}}{2}\) \(\overrightarrow{\mathrm{AD}} =\hat{i}+2 \hat{j}+3 \hat{k}\) \(\text { Length of Median } =\sqrt{1^2+2^2+3^2}\) \(=\sqrt{1+4+9}=\sqrt{14}\)
Karnataka CET-2020
Vector Algebra
87738
If \(\vec{a}=3, \vec{b}=4, \vec{c}=5\) each one of \(\vec{a}, \vec{b} \& \vec{c}\) is perpendicular to the sum of the remaining then \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}]\) is equal to
87736
If three vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}-\mathbf{3} \hat{\mathbf{k}}\) and \(3 \hat{i}+\lambda \hat{i}+5 \hat{k}\) are coplanar, then the value of \(\lambda\) is
87737
The two vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+3 \hat{j}+5 \hat{k}\) represents the two sides \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) respectively of a \(\triangle \mathrm{ABC}\). The length of the median though \(A\) is
1 14
2 7
3 \(\sqrt{14}\)
4 \(\frac{\sqrt{14}}{2}\)
Explanation:
(C) : Given, \(\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) We know that, \(\text { Median } =\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+3 \hat{j}+5 \hat{k})}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{2 \hat{i}+4 \hat{j}+6 \hat{k}}{2}\) \(\overrightarrow{\mathrm{AD}} =\hat{i}+2 \hat{j}+3 \hat{k}\) \(\text { Length of Median } =\sqrt{1^2+2^2+3^2}\) \(=\sqrt{1+4+9}=\sqrt{14}\)
Karnataka CET-2020
Vector Algebra
87738
If \(\vec{a}=3, \vec{b}=4, \vec{c}=5\) each one of \(\vec{a}, \vec{b} \& \vec{c}\) is perpendicular to the sum of the remaining then \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}]\) is equal to
87736
If three vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}-\mathbf{3} \hat{\mathbf{k}}\) and \(3 \hat{i}+\lambda \hat{i}+5 \hat{k}\) are coplanar, then the value of \(\lambda\) is
87737
The two vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+3 \hat{j}+5 \hat{k}\) represents the two sides \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) respectively of a \(\triangle \mathrm{ABC}\). The length of the median though \(A\) is
1 14
2 7
3 \(\sqrt{14}\)
4 \(\frac{\sqrt{14}}{2}\)
Explanation:
(C) : Given, \(\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) We know that, \(\text { Median } =\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+3 \hat{j}+5 \hat{k})}{2}\) \(\overrightarrow{\mathrm{AD}} =\frac{2 \hat{i}+4 \hat{j}+6 \hat{k}}{2}\) \(\overrightarrow{\mathrm{AD}} =\hat{i}+2 \hat{j}+3 \hat{k}\) \(\text { Length of Median } =\sqrt{1^2+2^2+3^2}\) \(=\sqrt{1+4+9}=\sqrt{14}\)
Karnataka CET-2020
Vector Algebra
87738
If \(\vec{a}=3, \vec{b}=4, \vec{c}=5\) each one of \(\vec{a}, \vec{b} \& \vec{c}\) is perpendicular to the sum of the remaining then \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}]\) is equal to
