Algebra Questions - JEE

A vector $\vec{C}$ directed along internal bisector of angle between vectors $\vec{A} = 7\vec{i} - 4\vec{j} - 4\vec{k}$ and $\vec{B} = -2\vec{i} - \vec{j} + 2\vec{k}$ with $|\vec{C}| = 5\sqrt{6}$ is

The sine of the angle between the vectors a = 3i + j + k and b = 2i - 2j + k is

In a parallelogram ABCD, $\vec{AB} = \vec{i} + \vec{j} + \vec{k}$ and diagonal $\vec{AC} = \vec{i} - \vec{j} + \vec{k}$ and area of parallelogram is 8 sq units, then $\angle BAC$ is equal to

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v}\cdot\vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^2$ is equal to:

Let $\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{v}\cdot\vec{w} = 2$ and $\vec{v}\times\vec{w} = \vec{u} + \lambda\vec{v}$. Then $\vec{u}\cdot\vec{w}$ is equal to

Let $\vec{\alpha} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{\beta} = \hat{i} + 2\hat{j} - 4\hat{k}$. Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$, then the value of $5\vec{\beta}_2\cdot(\hat{i}+\hat{j}+\hat{k})$ is

Let $\vec{a} = \hat{i} + 2\hat{j} + \lambda\hat{k}$, $\vec{b} = 3\hat{i} - 5\hat{j} - \lambda\hat{k}$, $\vec{a}\cdot\vec{c} = 7$, $2\vec{b}\cdot\vec{c} + 43 = 0$, $\vec{a}\times\vec{c} = \vec{b}\times\vec{c}$. Then $|\vec{a}\cdot\vec{b}|$ is equal to

Let $\lambda \in \mathbb{R}$, $\vec{a} = \lambda\hat{i}+2\hat{j}-3\hat{k}$, $\vec{b} = \hat{i}-\lambda\hat{j}+2\hat{k}$. If $((\vec{a}+\vec{b})\times(\vec{a}\times\vec{b}))\times(\vec{a}-\vec{b}) = 8\hat{i}-40\hat{j}-24\hat{k}$, then $|\lambda(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})|^2$ is equal to

Magnitude of the resultant displacement is given by :

Two adjacent sides of a parallelogram ABCD are given by $\overrightarrow{AB} = 2\vec{i} + 10\vec{j} + 11\vec{k}$ and $\overrightarrow{AD} = -\vec{i} + 2\vec{j} + 2\vec{k}$. The side AD is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes $\overrightarrow{AD'}$. If $\overrightarrow{AD'}$ makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by

Ex. 46 Statement I: If $|\mathbf{a} + \mathbf{b}| = |\mathbf{a} - \mathbf{b}|$, then $\mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other.Statement II: If the diagonals of a parallelogram are equal in magnitude, then the parallelogram is a rectangle.

Let a = $2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and b = $\mathbf{i} + \mathbf{j}$. If c is a vector such that $\mathbf{a} \cdot \mathbf{c} = |\mathbf{c}|$, $|\mathbf{c} - \mathbf{a}| = 2\sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and c is 30°, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|$ is equal to

The non-zero vectors a, b and c are related by a = 8b and c = -7b. Find the angle between a and c.

If vectors $\vec{a} = \frac{\vec{i} + \vec{j}}{\sqrt{2}}, \vec{b} = \frac{-\vec{i} + \vec{j}}{\sqrt{2}}$ and $\vec{c} = \vec{k}$ then the value of $(\vec{r} \cdot \vec{a})^2 + (\vec{r} \cdot \vec{b})^2 + (\vec{r} \cdot \vec{c})^2$ is equal to:

Ex. 56 In the regular pentagon ABCDE with the given position vectors, AD divides EC in the ratio

If the volume of parallelepiped formed by the vectors a, b, c as three coterminous edges is 27 cu units, then the volume of the parallelepiped with $\alpha = a + 2b - c$, $\beta = a - b$ and $\gamma = a - b - c$ as three coterminous edges is

The position vectors of the points A, B and C are i + 2j - k, i + j + k, and i + 3j + 2k, respectively. If A is [incomplete question]

If O be the circumcentre and O' be the orthocentre of the $\triangle ABC$, then $\vec{O'A} + \vec{O'B} + \vec{O'C}$ is equal to

A line segment has length 63 and direction ratios are 3, -2 and 6. Find the components of the line vector.

The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is

$(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$ is $\vec{a}$ and $\vec{c}$ are:

Let the cosine of angle between the vectors p and q be $λ$ such that $2\mathbf{p} + \mathbf{q} = \mathbf{i} + \mathbf{j}$ and $\mathbf{p} + 2\mathbf{q} = \mathbf{i} - \mathbf{j}$, then $λ$ is equal to

The position vector of the points which divides internally in the ratio 2 : 3 the join of the points 2a - 3b and 3a - 2b, is

Consider points $A$, $B$, $C$ and $D$ with position vectors $7\hat{i} - 4\hat{j} + 7\hat{k}$, $\hat{i} - 6\hat{j} + 10\hat{k}$, $-\hat{i} - 3\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + \hat{k}$, respectively. Then $ABCD$ is a

Let $a$, $b$, $c$ be three vectors such that $[a b c] = 2$. If $r = l(b \times c) + m(c \times a) + n(a \times b)$ is perpendicular to $a + b + c$, then the value of $(l + m + n)$ is

Let $\vec{a} = \vec{i} + 2\vec{j} + \vec{k}$, $\vec{b} = \vec{i} - \vec{j} + \vec{k}$, $\vec{c} = \vec{i} + \vec{j} - \vec{k}$. A vector coplanar to $\vec{a}$ and $\vec{b}$ has a projection along $\vec{c}$ of magnitude $\frac{1}{3}$. Then the vector is:

Given that $\vec{a}$ is perpendicular to $\vec{b}$ and p is a non-zero scalar, if $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$ then $\vec{r} = $:

If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are on a circle of radius $R$ whose centre is at origin and $\vec{c} - \vec{a}$ is perpendicular to $\vec{d} - \vec{b}$, then $|\vec{d} - \vec{a}|^2 + |\vec{b} - \vec{c}|^2$ (AC is diameter)

Let two non-collinear unit vectors $\vec{a}$ and $\vec{b}$ form an acute angle. A point P moves, so that at any time $t$ the position vector $\overrightarrow{OP}$ (where O is the origin) is given by $\overrightarrow{OP} = \vec{a}\cos t + \vec{b}\sin t$. When P is farthest from origin O, let M be the length of $\overrightarrow{OP}$ and $\vec{u}$ be the unit vector along $\overrightarrow{OP}$. Then find $\vec{u}$ and $M$.

If the position vectors of the points A and B are i + 3j - k and 3i - j - 3k, then what will be the position vector of the mid-point of AB?

The vector $\vec{OP} = 5\vec{i} + 12\vec{j} + 13\vec{k}$ turns through an angle of $\frac{\pi}{2}$ about $O$ passing through the positive side of $\vec{j}$ axis an iff way. The vector in the new position is:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three non-parallel unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2}\vec{b}$, the angle between $\vec{b}$ and $\vec{c}$ is:

Example 32. The length of longer diagonal of the parallelogram constructed on $5\mathbf{a} + 2\mathbf{b}$ and $\mathbf{a} - 3\mathbf{b}$, when it is given that $|\mathbf{a}| = 2\sqrt{2}$, $|\mathbf{b}| = 3$ and angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{\pi}{4}$, is

Let r, a, b and c be four non-zero vectors such that r · a = 0, |r × b| = |r| |b|, |r × c| = |r| |c|, then [a b c] is

Given $\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$. Vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear. Find $\lambda$.

What should be added to vector a = 3i + 4j - 2k to get its resultant a unit vector i?

Let $a, b, c$ denote the lengths of the sides of a triangle such that $(a-b)\vec{u} + (b-c)\vec{v} + (c-a)(\vec{u} \times \vec{v}) = 0$ for any two non-collinear vectors $\vec{u}$ and $\vec{v}$, then the triangle is

If O is origin and C is the mid-point of A(2, -1) and B(-4, 3), then the value of OC is

Let $OPQR$ is a tetrahedon such that O is origin and $\vec{p}, \vec{q}, \vec{r}$ are position vectors of P, Q, R respectively and $\alpha$ is the angle which OP makes with face PQR then:

Which of the following statements is/are correct? (a) If $\mathbf{n} \cdot \mathbf{a} = 0, \mathbf{n} \cdot \mathbf{b} = 0$ and $\mathbf{n} \cdot \mathbf{c} = 0$ for some non-zero vector $\mathbf{n}$, then $[\mathbf{a} \mathbf{b} \mathbf{c}] = 0$

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be distinct non-negative numbers. If the vectors $a\vec{i} + a\vec{j} + c\vec{k}$, $\vec{i} + \vec{k}$ and $c\vec{i} + c\vec{j} + b\vec{k}$ lie in a plane, then $c$ is

If the interior and exterior bisectors of the angle $A$ of a triangle $ABC$ meet the side $BC$ at $D$ and $E$, then :

Let OABCD be a pentagon in which the sides OA and CB are parallel and the sides OD and AB are parallel. Also, OA : CB = 2 : 1 and OD : AB = 1 : 3. The ratio $\frac{AX}{XD}$ is

The direction cosines of vector a = 3i + 4j + 5k in the direction of positive axis of X, is

Ex. 64: Find the least positive integral value of x for which the angle between vectors $\vec{a} = x\vec{i} - 3\vec{j} - \vec{k}$ and $\vec{b} = 2x\vec{i} + x\vec{j} - \vec{k}$ is acute.

The resultant displacement of the ant after first two steps is:

The vector, directed along the internal bisector of the angle between the vectors $\vec{a} = 7\vec{i} - 4\vec{j} - 4\vec{k}$ & $\vec{b} = -2\vec{i} - \vec{j} + 2\vec{k}$ with $|\vec{c}| = 5\sqrt{6}$ is:

Ex. 54 The ratio $PQ : DB$ is equal to

Ex. 106: Find AN + BP + CM where points M, N and P are taken on sides AB, BC and CA respectively, such that \frac{AM}{AB} = \frac{BN}{BC} = \frac{CP}{CA} = \alpha, with A at the origin, B and C having position vectors b and c respectively.

OP · OQ, where O is the origin (P is on the circle and curve, Q is inside the circle with integer abscissa)

Algebra Questions (626)