JEE Main Vector Algebra Practice Questions With Solutions

Between the following two statements:

Statement I : Let $\overrightarrow{a}=\hat{i}+2\hat{j}-3\hat{k}$\overset{\rightarrow}{a} = \hat{i} + 2 \hat{j} - 3 \hat{k} and $\overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k}$\overset{\rightarrow}{b} = 2 \hat{i} + \hat{j} - \hat{k}. Then the vector $\overrightarrow{r}$\overset{\rightarrow}{r} satisfying $\overrightarrow{a}\times \overrightarrow{r}=\overrightarrow{a}\times \overrightarrow{b}$\overset{\rightarrow}{a} \times \overset{\rightarrow}{r} = \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} and $\overrightarrow{a}\cdot \overrightarrow{r}=0$\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{r} = 0 is of magnitude $\sqrt{10}$\sqrt{10}.

Statement II : In a triangle $ABC,\cos 2A+\cos 2B+\cos 2C\ge -\frac{3}{2}$A B C , cos ⁡ 2 A + cos ⁡ 2 B + cos ⁡ 2 C \geq - \frac{3}{2}.

Let $\overrightarrow{a}=2\hat{i}+\alpha \hat{j}+\hat{k},\overrightarrow{b}=-\hat{i}+\hat{k},\overrightarrow{c}=\beta \hat{j}-\hat{k}$\overset{\rightarrow}{a} = 2 \hat{i} + \alpha \hat{j} + \hat{k} , \overset{\rightarrow}{b} = - \hat{i} + \hat{k} , \overset{\rightarrow}{c} = \beta \hat{j} - \hat{k}, where $\alpha$\alpha and $\beta$\beta are integers and $\alpha \beta =-6$\alpha \beta = - 6. Let the values of the ordered pair $(\alpha ,\beta )$\left(\right. \alpha , \beta \left.\right), for which the area of the parallelogram of diagonals $\overrightarrow{a}+\overrightarrow{b}$\overset{\rightarrow}{a} + \overset{\rightarrow}{b} and $\overrightarrow{b}+\overrightarrow{c}$\overset{\rightarrow}{b} + \overset{\rightarrow}{c} is $\frac{\sqrt{21}}{2}$\frac{\sqrt{21}}{2}, be $({\alpha }_1,{\beta }_1)$\left(\right. \alpha_{1} , \beta_{1} \left.\right) and $({\alpha }_2,{\beta }_2)$\left(\right. \alpha_{2} , \beta_{2} \left.\right). Then ${\alpha }_1^2+{\beta }_1^2-{\alpha }_2{\beta }_2$\alpha_{1}^{2} + \beta_{1}^{2} - \alpha_{2} \beta_{2} is equal to

Let three vectors ,$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4\hat{j}+2\hat{k},\overrightarrow{\mathrm{b}}=5\hat{i}+3\hat{j}+4\hat{k},\overrightarrow{\mathrm{c}}=x\hat{i}+y\hat{j}+z\hat{k}$\overset{\rightarrow}{a} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k} , \overset{\rightarrow}{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} , \overset{\rightarrow}{c} = x \hat{i} + y \hat{j} + z \hat{k} form a triangle such that $\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{b}$\overset{\rightarrow}{c} = \overset{\rightarrow}{a} - \overset{\rightarrow}{b} and the area of the triangle is $5\sqrt{6}$5 \sqrt{6}. If $\alpha$\alpha is a positive real number, then $|\overrightarrow{c}{|}^2$\left|\right. \overset{\rightarrow}{c} \left|\right.^{2} is equal to:

Let $\overrightarrow{OA}=2\overrightarrow{a},\overrightarrow{OB}=6\overrightarrow{a}+5\overrightarrow{b}$\overset{\rightarrow}{O A} = 2 \overset{\rightarrow}{a} , \overset{\rightarrow}{O B} = 6 \overset{\rightarrow}{a} + 5 \overset{\rightarrow}{b} and $\overrightarrow{OC}=3\overrightarrow{b}$\overset{\rightarrow}{O C} = 3 \overset{\rightarrow}{b}, where $O$O is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$\overset{\rightarrow}{O A} and $\overrightarrow{OC}$\overset{\rightarrow}{O C} is 15 sq. units, then the area (in sq. units) of the quadrilateral $OABC$O A B C is equal to:

Let $\overrightarrow{\mathrm{a}}=4\hat{i}-\hat{j}+\hat{k},\overrightarrow{\mathrm{b}}=11\hat{i}-\hat{j}+\hat{k}$\overset{\rightarrow}{a} = 4 \hat{i} - \hat{j} + \hat{k} , \overset{\rightarrow}{b} = 11 \hat{i} - \hat{j} + \hat{k} and $\overrightarrow{\mathrm{c}}$\overset{\rightarrow}{c} be a vector such that $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}}\times (-2\overrightarrow{\mathrm{a}}+3\overrightarrow{\mathrm{b}})$\left(\right. \overset{\rightarrow}{a} + \overset{\rightarrow}{b} \left.\right) \times \overset{\rightarrow}{c} = \overset{\rightarrow}{c} \times \left(\right. - 2 \overset{\rightarrow}{a} + 3 \overset{\rightarrow}{b} \left.\right).If $(2\overrightarrow{a}+3\overrightarrow{b})\cdot \overrightarrow{c}=1670$\left(\right. 2 \overset{\rightarrow}{a} + 3 \overset{\rightarrow}{b} \left.\right) \cdot \overset{\rightarrow}{c} = 1670, then $|\overrightarrow{c}{|}^2$\left|\right. \overset{\rightarrow}{c} \left|\right.^{2} is equal to:

Let $\overrightarrow{\mathrm{a}}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{\mathrm{b}}=2\hat{i}+3\hat{j}-5\hat{k}$\overset{\rightarrow}{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} , \overset{\rightarrow}{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k} and $\overrightarrow{\mathrm{c}}=3\hat{i}-\hat{j}+\lambda \hat{k}$\overset{\rightarrow}{c} = 3 \hat{i} - \hat{j} + \lambda \hat{k} be three vectors. Let $\overrightarrow{\mathrm{r}}$\overset{\rightarrow}{r} be a unit vector along $\overrightarrow{b}+\overrightarrow{c}$\overset{\rightarrow}{b} + \overset{\rightarrow}{c}. If $\overrightarrow{r}\cdot \overrightarrow{a}=3$\overset{\rightarrow}{r} \cdot \overset{\rightarrow}{a} = 3, then $3\lambda$3 \lambda is equal to:

The set of all $\alpha$\alpha, for which the vectors $\overrightarrow{a}=\alpha t\hat{i}+6\hat{j}-3\hat{k}$\overset{\rightarrow}{a} = \alpha t \hat{i} + 6 \hat{j} - 3 \hat{k} and $\overrightarrow{b}=t\hat{i}-2\hat{j}-2\alpha t\hat{k}$\overset{\rightarrow}{b} = t \hat{i} - 2 \hat{j} - 2 \alpha t \hat{k} are inclined at an obtuse angle for all $t\in \mathbb{R}$t \in \mathbb{R}, is

Let $\overrightarrow{a}=2\hat{i}+\hat{j}-\hat{k},\overrightarrow{b}=((\overrightarrow{a}\times (\hat{i}+\hat{j}))\times \hat{i})\times \hat{i}$\overset{\rightarrow}{a} = 2 \hat{i} + \hat{j} - \hat{k} , \overset{\rightarrow}{b} = \left(\right. \left(\right. \overset{\rightarrow}{a} \times \left(\right. \hat{i} + \hat{j} \left.\right) \left.\right) \times \hat{i} \left.\right) \times \hat{i}. Then the square of the projection of $\overrightarrow{a}$\overset{\rightarrow}{a} on $\overrightarrow{b}$\overset{\rightarrow}{b} is:

Let $\overrightarrow{\mathrm{a}}=6\hat{i}+\hat{j}-\hat{k}$\overset{\rightarrow}{a} = 6 \hat{i} + \hat{j} - \hat{k} and $\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$\overset{\rightarrow}{b} = \hat{i} + \hat{j}. If $\overrightarrow{\mathrm{c}}$\overset{\rightarrow}{c} is a is vector such that $|\overrightarrow{\mathrm{c}}|\ge 6,\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2\sqrt{2}$\left|\right. \overset{\rightarrow}{c} \left|\right. \geq 6 , \overset{\rightarrow}{a} \cdot \overset{\rightarrow}{c} = 6 \left|\right. \overset{\rightarrow}{c} \left|\right. , \left|\right. \overset{\rightarrow}{c} - \overset{\rightarrow}{a} \left|\right. = 2 \sqrt{2} and the angle between $\overrightarrow{a}\times \overrightarrow{b}$\overset{\rightarrow}{a} \times \overset{\rightarrow}{b} and $\overrightarrow{c}$\overset{\rightarrow}{c} is ${60}^{\circ }$60^{\circ}, then $|(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|$\left|\right. \left(\right. \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} \left.\right) \times \overset{\rightarrow}{c} \left|\right. is equal to:

Let $\overrightarrow{a}=2\hat{i}+5\hat{j}-\hat{k},\overrightarrow{b}=2\hat{i}-2\hat{j}+2\hat{k}$\overset{\rightarrow}{a} = 2 \hat{i} + 5 \hat{j} - \hat{k} , \overset{\rightarrow}{b} = 2 \hat{i} - 2 \hat{j} + 2 \hat{k} and $\overrightarrow{c}$\overset{\rightarrow}{c} be three vectors such that $(\overrightarrow{c}+\hat{i})\times (\overrightarrow{a}+\overrightarrow{b}+\hat{i})=\overrightarrow{a}\times (\overrightarrow{c}+\hat{i})$\left(\right. \overset{\rightarrow}{c} + \hat{i} \left.\right) \times \left(\right. \overset{\rightarrow}{a} + \overset{\rightarrow}{b} + \hat{i} \left.\right) = \overset{\rightarrow}{a} \times \left(\right. \overset{\rightarrow}{c} + \hat{i} \left.\right). If $\overrightarrow{a}\cdot \overrightarrow{c}=-29$\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{c} = - 29, then $\overrightarrow{c}\cdot (-2\hat{i}+\hat{j}+\hat{k})$\overset{\rightarrow}{c} \cdot \left(\right. - 2 \hat{i} + \hat{j} + \hat{k} \left.\right) is equal to: