Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

The scalar triple product $a \cdot (b \times c)$ is equivalent to the dot product $(a \times b) \cdot c$. Given the relation $a \times c = b$, we can use the vector triple product identity $a \times (a \times c) = (a \cdot c)a - (a \cdot a)c$ to express the cross product $a \times b$ in terms of $a$ and $c$. Substituting the given values $a = \hat{i} + \hat{j} + \hat{k}$, $a \cdot c = 3$, and the magnitude squared $|a|^2 = 1^2 + 1^2 + 1^2 = 3$ into this identity results in $a \times b = 3a - 3c$, which rearranges to $3c = 3a - (a \times b)$.

Inserting this expression for $3c$ into our original scalar triple product yields $\frac{1}{3} (a \times b) \cdot (3a - (a \times b))$. Because the cross product $a \times b$ is by definition orthogonal to vector $a$, the term $(a \times b) \cdot 3a$ becomes zero, leaving only $-\frac{1}{3} |a \times b|^2$. Computing the cross product $a \times b$ yields $-2\hat{i} + \hat{j} + \hat{k}$, and the squared magnitude of this vector is $(-2)^2 + 1^2 + 1^2 = 6$. Applying this to the simplified expression gives $-\frac{1}{3} \times 6 = -2$.

Simplify the nested vector expression by utilizing the algebraic properties of cross products and the vector triple product identity. Given that the cross product of any vector with itself is the zero vector, the term $((\vec{a} - \vec{b}) \times \vec{b})$ simplifies to $(\vec{a} \times \vec{b}) - (\vec{b} \times \vec{b})$, which is simply $\vec{a} \times \vec{b}$. This reduces the expression to $(\vec{a} \times (\vec{a} \times \vec{b})) \times \vec{b}$. Applying the vector triple product rule, where $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$, the term $\vec{a} \times (\vec{a} \times \vec{b})$ expands to $\vec{a}(\vec{a} \cdot \vec{b}) - \vec{b}(\vec{a} \cdot \vec{a})$. Substituting this back into the overall expression results in $(\vec{a}(\vec{a} \cdot \vec{b}) - \vec{b}(\vec{a} \cdot \vec{a})) \times \vec{b}$, which simplifies further to $(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b}) - (\vec{a} \cdot \vec{a})(\vec{b} \times \vec{b})$. Since $\vec{b} \times \vec{b} = \vec{0}$, the complex vector portion reduces entirely to $(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b})$.

With the vectors defined as $\vec{a} = \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} + 3\hat{k}$, the dot product $\vec{a} \cdot \vec{b}$ is $(1)(-1) + (1)(2) + (2)(3) = -1 + 2 + 6 = 7$. The cross product $\vec{a} \times \vec{b}$ is computed via the determinant of the basis vectors $\hat{i}, \hat{j}, \hat{k}$ with the coefficients of $\vec{a}$ and $\vec{b}$, resulting in $-\hat{i} - 5\hat{j} + 3\hat{k}$. Multiplying this vector by the scalar dot product 7 yields $7(-\hat{i} - 5\hat{j} + 3\hat{k})$. Calculating the sum $\vec{a} + \vec{b}$ gives $3\hat{j} + 5\hat{k}$. The final cross product of $(3\hat{j} + 5\hat{k})$ and $7(-\hat{i} - 5\hat{j} + 3\hat{k})$ is determined by evaluating $7$ times the determinant with rows $\hat{i}, \hat{j}, \hat{k}$, $0, 3, 5$, and $-1, -5, 3$. This calculation yields $34\hat{i} - 5\hat{j} + 3\hat{k}$, which results in $7(34\hat{i} - 5\hat{j} + 3\hat{k})$.

When three vectors are coplanar, their scalar triple product must be zero, which means the determinant of the matrix formed by their components must equal zero. The condition of coplanarity results in the following equation:

$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$

Expanding this determinant along the first row leads to the expression $a(0 - c) - a(b - c) + c(c - 0) = 0$. Simplifying this expression yields $-ac - ab + ac + c^2 = 0$, which reduces directly to $c^2 - ab = 0$. Given that $c$ is a positive number, it follows that $c = \sqrt{ab}$.

The shortest distance between two skew lines defined by $\vec{r} = \vec{a} + \lambda\vec{b}$ and $\vec{r} = \vec{c} + \mu\vec{d}$ is found by projecting the vector connecting any two points on the lines onto the common perpendicular unit vector. This geometric principle is represented by the formula
$L = \left| (\vec{a} - \vec{c}) \cdot \frac{\vec{b} \times \vec{d}}{|\vec{b} \times \vec{d}|} \right|$
where the difference between the position vectors is $\vec{a} - \vec{c} = (\alpha + 4)\hat{i} + 2\hat{j} + 3\hat{k}$ and the unit vector normal to the lines is $\frac{2\hat{i} + 2\hat{j} + \hat{k}}{3}$.

Substituting these expressions into the distance formula and equating the result to 9 gives
$\frac{|(\alpha + 4)(2) + (2)(2) + (3)(1)|}{3} = 9$
which expands to the relationship $2(\alpha + 4) + 4 + 3 = 27$. Simplifying this equation leads to $2\alpha + 15 = 27$, confirming that $\alpha = 6$.

The magnitude of the cross product is defined by $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta$. Substituting the given values, $8 = 2 \times 5 \sin \theta$, we find that $\sin \theta = \frac{4}{5}$. By applying the identity $\cos^2 \theta = 1 - \sin^2 \theta$, we get $\cos^2 \theta = 1 - (\frac{4}{5})^2 = \frac{9}{25}$, which means $|\cos \theta| = \frac{3}{5}$. The magnitude of the dot product $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos \theta|$ then becomes $2 \times 5 \times \frac{3}{5} = 6$.

The initial vector $\sqrt{3} \hat{i} + \hat{j}$ can be defined by its magnitude and the angle it makes with the positive x-axis. Calculating the magnitude using the Pythagorean theorem gives $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$, and the inclination angle is $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^{\circ}$. Rotating this vector by $45^{\circ}$ counterclockwise increases the inclination angle to $30^{\circ} + 45^{\circ} = 75^{\circ}$ while maintaining the same magnitude of 2. Consequently, the coordinates $(\alpha, \beta)$ of the rotated vector are given by $(2 \cos 75^{\circ}, 2 \sin 75^{\circ})$.

The triangle formed by the vertices $(0,0)$, $(0, \beta)$, and $(\alpha, \beta)$ is a right-angled triangle with a base of length $\alpha$ and a height of length $\beta$. The area of this triangle is expressed as $\frac{1}{2} \alpha \beta$. Substituting the coordinate values leads to $\frac{1}{2} (2 \cos 75^{\circ}) (2 \sin 75^{\circ})$, which simplifies to the trigonometric identity $2 \sin 75^{\circ} \cos 75^{\circ}$. Using the double-angle formula $\sin 2\theta = 2 \sin \theta \cos \theta$, this expression becomes $\sin 150^{\circ}$, which results in an area of 0.5.

Any vector $\vec{x}$ lying in the plane of $\vec{a} = 2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}-\hat{k}$ can be expressed as a linear combination $\vec{x} = \lambda\vec{a} + \mu\vec{b}$. Because $\vec{x}$ is orthogonal to the vector $3\hat{i}+2\hat{j}-\hat{k}$, their dot product equals zero, which simplifies to the linear relation $3\lambda + 8\mu = 0$. The projection of $\vec{x}$ onto $\vec{a}$ is given by $\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|}$, and setting this equal to $\frac{17\sqrt{6}}{2}$ with $|\vec{a}| = \sqrt{6}$ implies that the dot product $\vec{x} \cdot \vec{a}$ must be $51$. Substituting the linear combination for $\vec{x}$ into this dot product leads to the second equation $6\lambda - \mu = 51$. Solving these two linear equations simultaneously yields the coefficients $\lambda = 8$ and $\mu = -3$. Substituting these back into the expression for $\vec{x}$ gives $8(2\hat{i}-\hat{j}+\hat{k}) - 3(\hat{i}+2\hat{j}-\hat{k})$, which simplifies to the vector $\vec{x} = 13\hat{i}-14\hat{j}+11\hat{k}$. Calculating the squared magnitude produces $|\vec{x}|^{2} = 13^{2} + (-14)^{2} + 11^{2} = 486$.

The area of a parallelogram defined by two vectors $\mathbf{a}$ and $\mathbf{b}$ corresponds to the magnitude of their cross product, $|\mathbf{a} \times \mathbf{b}|$. By evaluating the determinant with the given vectors $\mathbf{a} = \hat{i} + \alpha \hat{j} + 3\hat{k}$ and $\mathbf{b} = 3\hat{i} - \alpha \hat{j} + \hat{k}$, the cross product results in the vector $4\alpha\hat{i} + 8\hat{j} - 4\alpha\hat{k}$. The magnitude of this resulting vector is $\sqrt{(4\alpha)^2 + 8^2 + (-4\alpha)^2}$, which simplifies to $\sqrt{32\alpha^2 + 64}$.

Equating this magnitude to the provided area of $8\sqrt{3}$ and squaring both sides gives the equation $32\alpha^2 + 64 = 192$. Subtracting 64 from both sides results in $32\alpha^2 = 128$, which yields $\alpha^2 = 4$. Calculating the dot product $\mathbf{a} \cdot \mathbf{b}$ by multiplying corresponding components gives the expression $3 - \alpha^2 + 3$, which simplifies to $6 - \alpha^2$. Substituting the derived value of $\alpha^2$ into this expression results in 2.

Three vectors are coplanar if and only if their scalar triple product is zero, a condition which implies that the determinant of the matrix formed by their coefficients must vanish. Constructing the determinant from the provided vectors and applying the row transformations $R_3 \rightarrow R_3 - R_2$ and $R_1 \rightarrow R_1 - R_2$ simplifies the expression to

Evaluating this determinant by expanding along the first row yields $(a+1)(2b) - (a+c)(2b+1) - c(-2b) = 0$. Distributing these terms results in $2ab + 2b - (2ab + a + 2bc + c) + 2bc = 0$, which further simplifies to $2ab + 2b - 2ab - a - 2bc - c + 2bc = 0$. Combining like terms leaves the linear equation $2b - a - c = 0$, which rearranges to $2b = a + c$.

The key to simplifying the expression $\vec{a} = \vec{b} \times (\vec{b} \times \vec{c})$ lies in the vector triple product identity, which states that $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$. Applying this rule to the given setup yields $\vec{a} = (\vec{b} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{b})\vec{c}$. Since the magnitudes are $|\vec{b}| = 1$ and $|\vec{c}| = 2$, and the dot product $\vec{b} \cdot \vec{c}$ equals $2 \cos \theta$, the equation becomes $\vec{a} = 2 \cos \theta \vec{b} - \vec{c}$.

Squaring both sides of this vector equation relates the result back to the known magnitude $|\vec{a}| = \sqrt{2}$. This operation yields $|\vec{a}|^2 = |2 \cos \theta \vec{b} - \vec{c}|^2$, which expands to $(2 \cos \theta)^2 |\vec{b}|^2 + |\vec{c}|^2 - 2(2 \cos \theta)(\vec{b} \cdot \vec{c})$. Substituting the known values $2 = 4 \cos^2 \theta(1) + 4 - 4 \cos \theta(2 \cos \theta)$ leads to $2 = 4 \cos^2 \theta + 4 - 8 \cos^2 \theta$, which simplifies to $2 = 4 - 4 \cos^2 \theta$. Rearranging produces $4 \cos^2 \theta = 2$, or $\cos^2 \theta = 1/2$. Since $\theta$ is acute, $\cos \theta = 1/\sqrt{2}$, implying $\theta = \pi/4$. With $\tan (\pi/4) = 1$, the final value $1 + \tan \theta$ is 2.

Given, angle between A and B, θ = 120° The angle between A and - B, α = 180° − θ = 180° − 120° = 60° Now, from diagram the resultant vector of A and −B will be A − B and the angle between A − B and A is denoted by β. So, angle between A and −B is calculated as, ϕ = 180°− α = 180°− 60° = 120° Now, from parallelogram law of vector addition angle β can becalculated as follows $\tan \beta =\frac{-\text{B} \sin \phi }{-\text{A}+(-\text{B} \cos \phi )}$ $=\frac{-\text{B} \sin 120°}{-\text{A}-\text{B} \cos 120°}={-\text{B} \frac{\\sqrt{3}}{2}}/{-\text{A}+\frac{\text{B}}{2}}$ $=\frac{\\sqrt{3} \text{B}}{2 \text{A}\-\text{B}}$ $\beta =\tan ^{-1}(\frac{\\sqrt{3} \text{B}}{2 \text{A} - \text{B}})$ Hence, the angle between vector A and (A - B) is $\tan ^{-1}(\frac{\\sqrt{3} \text{B}}{2 \text{A} - \text{B}})$.

The projection of vector $\overrightarrow{BA}$ onto $\overrightarrow{BC}$ is determined by the formula $|\overrightarrow{BA}| \cos(\angle ABC)$. Applying the Law of Cosines to the triangle with side lengths $|\overrightarrow{BC}|=3$, $|\overrightarrow{CA}|=5$, and $|\overrightarrow{BA}|=7$ yields $\cos(\angle ABC) = \frac{7^2 + 3^2 - 5^2}{2 \times 7 \times 3}$. Multiplying the magnitude $|\overrightarrow{BA}|=7$ by this cosine value simplifies to $7 \times \frac{33}{42}$, which results in $\frac{11}{2}$.

(a) ${\vec{C}}={\vec{A}}+{\vec{B}}$ Option (iv) (b) ${\vec{A}}={\vec{B}}+{\vec{C}}={\vec{C}}+{\vec{B}}$ Option (iii) (c) ${\vec{B}}={{A}}^{\to }+{\vec{C}}$ Option (i) (d) ${\vec{A}}+{\vec{B}}+{{C}}^{\to }=0$ Option (ii)

Any vector $\vec{r}$ that is perpendicular to both $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$ must lie along the direction of their cross product, as this product represents the vector normal to the plane containing both input vectors. Given $\vec{p}=2\hat{i}+3\hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2\hat{j}+\hat{k}$, we first find $(\vec{p}+\vec{q})=3\hat{i}+5\hat{j}+2\hat{k}$ and $(\vec{p}-\vec{q})=\hat{i}+\hat{j}$. The cross product $(\vec{p}+\vec{q}) \times (\vec{p}-\vec{q})$ is calculated using the determinant

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 1 & 1 & 0 \end{vmatrix} = -2\hat{i} + 2\hat{j} - 2\hat{k}$

The magnitude of this cross product is $\sqrt{(-2)^{2+}2^{2+}(-2)^2} = \sqrt{12} = 2\sqrt{3}$. To find $\vec{r}$ given $|\vec{r}|=\sqrt{3}$, we multiply the normalized unit vector by the required magnitude, resulting in $\vec{r} = \pm \sqrt{3} \left( \frac{-2\hat{i}+2\hat{j}-2\hat{k}}{2\sqrt{3}} \right)$, which simplifies to $\vec{r} = \pm(-\hat{i} + \hat{j} - \hat{k})$. With $\vec{r} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$, the absolute values of the components are $|\alpha|=1$, $|\beta|=1$, and $|\gamma|=1$. Summing these values yields $|\alpha|+|\beta|+|\gamma| = 3$.

To solve this problem, we rely on the orthogonality condition for perpendicular vectors and the scalar triple product property for coplanar vectors. Since $\vec{OP}$ is perpendicular to $\vec{OQ}$, their dot product $(x)(-1) + (y)(2) + (-1)(3x) = -4x + 2y$ must equal zero, which simplifies to the relationship $y = 2x$. By calculating the vector $\vec{PQ} = \vec{OQ} - \vec{OP} = (-1-x)\hat{i} + (2-y)\hat{j} + (3x+1)\hat{k}$ and using the magnitude constraint $|\vec{PQ}| = \sqrt{20}$, we obtain the equation $(-1-x)^2 + (2-y)^2 + (3x+1)^2 = 20$. Substituting $y = 2x$ into this expression yields $14x^2 + 6 = 20$, which simplifies to $14x^2 = 14$, confirming that $x = 1$ and $y = 2$ given the constraint $x > 0$. Finally, the coplanarity of $\vec{OP}$, $\vec{OQ}$, and $\vec{OR}$ requires the determinant

to vanish. Solving this determinant results in $1(-14 - 3z) - 2(7 - 9) - 1(-z - 6) = 0$, which leads to $-14 - 3z + 4 + z + 6 = 0$ and finally $z = -2$. The requested sum $x^2 + y^2 + z^2$ evaluates to $1^2 + 2^2 + (-2)^2$, which equals $9$.

The geometric condition $r \times a = c \times a$ is equivalent to $(r - c) \times a = 0$, which signifies that the vector $r - c$ is parallel to $a$. This allows us to write $r - c = \lambda a$ for some scalar constant $\lambda$, meaning $r = \lambda a + c$. By applying the dot product with $b$ and using the constraint $r \cdot b = 0$, we obtain $(\lambda a + c) \cdot b = 0$, or $\lambda(a \cdot b) + c \cdot b = 0$. Using the provided vectors $a = \hat{i} + 2\hat{j} - \hat{k}$, $b = \hat{i} - \hat{j}$, and $c = \hat{i} - \hat{j} - \hat{k}$, we calculate the dot products $a \cdot b = 1 - 2 = -1$ and $c \cdot b = 1 + 1 = 2$. Substituting these values into our equation gives $-\lambda + 2 = 0$, which yields $\lambda = 2$.

With $\lambda$ determined, the vector becomes $r = 2a + c$. To find the value of $r \cdot a$, we compute the dot product of this expression with $a$, resulting in $r \cdot a = (2a + c) \cdot a = 2|a|^2 + c \cdot a$. We calculate the magnitude squared $|a|^2 = 1^2 + 2^2 + (-1)^2 = 6$ and the dot product $c \cdot a = (1)(1) + (-1)(2) + (-1)(-1) = 0$. Substituting these values into the expression gives $2(6) + 0$, leading to a final result of 12.

Two vectors are perpendicular if and only if their dot product equals zero, a property that allows us to establish a relationship between the unknowns $\alpha$ and $\beta$. Given $a = \hat{i} + 5 \hat{j} + \alpha \hat{k}$ and $b = \hat{i} + 3 \hat{j} + \beta \hat{k}$, the condition $a \cdot b = 0$ leads to the equation $1(1) + 5(3) + \alpha \beta = 0$, which simplifies to $16 + \alpha \beta = 0$, or $\alpha \beta = -16$.

The cross product $b \times c$ is calculated as the determinant of the matrix formed by the unit vectors and the components of $b$ and $c$, yielding $b \times c = \hat{i}(-9 - 2\beta) - \hat{j}(-3 + \beta) + \hat{k}(2 + 3)$, which simplifies to $(-9 - 2\beta)\hat{i} + (3 - \beta)\hat{j} + 5\hat{k}$. Since the magnitude $|b \times c| = 5 \sqrt{3}$, we square this result to get $|b \times c|^2 = 75$, resulting in the equation $(-9 - 2\beta)^2 + (3 - \beta)^2 + 25 = 75$. Expanding this leads to $\beta^2 + 6\beta + 8 = 0$, which factors into $(\beta + 2)(\beta + 4) = 0$, identifying the two possible values for $\beta$ as $-2$ and $-4$.

Substituting these values back into the relation $\alpha \beta = -16$ gives $\alpha = 8$ when $\beta = -2$, and $\alpha = 4$ when $\beta = -4$. The squared magnitude of vector $a$ is calculated as $|a|^2 = 1^2 + 5^2 + \alpha^2 = 26 + \alpha^2$. Evaluating this for both cases, $\alpha = 8$ results in $26 + 64 = 90$, while $\alpha = 4$ results in $26 + 16 = 42$. Comparing these results, the greatest value is 90.

The vector triple product identity $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$ provides the fundamental rule for expanding nested cross products. Since $a$ and $b$ are perpendicular, the dot product $a \cdot b = 0$, which simplifies the expression significantly by eliminating the component of the vector parallel to $a$.

Applying this identity to the innermost term $a \times (a \times b)$ yields $(a \cdot b)a - (a \cdot a)b$, which reduces to $-|a|^2 b$ because the first part of the expression vanishes. Replacing this result back into the full expression $a \times (a \times (a \times (a \times b)))$ creates $a \times (a \times (-|a|^2 b))$, which factors to $-|a|^2 (a \times (a \times b))$. Applying the identity once more to the remaining term leads to $-|a|^2 (-|a|^2 b)$, simplifying to a final result of $|a|^4 b$.

The vector triple product identity, which states that $\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w}$, provides the most direct path to simplifying the given expression. By applying this identity to the first term, $\mathbf{a} \times \{(\mathbf{r} - \mathbf{b}) \times \mathbf{a}\}$, one arrives at $|\mathbf{a}|^2(\mathbf{r} - \mathbf{b}) - (\mathbf{a} \cdot (\mathbf{r} - \mathbf{b}))\mathbf{a}$. Given that the vectors are mutually perpendicular, their dot products between distinct vectors are zero, which simplifies the expression to $|\mathbf{a}|^2(\mathbf{r} - \mathbf{b}) - (\mathbf{r} \cdot \mathbf{a})\mathbf{a}$. Carrying this expansion across all three terms of the original equation leads to the collective form $|\mathbf{a}|^2(3\mathbf{r} - (\mathbf{a} + \mathbf{b} + \mathbf{c})) - [(\mathbf{r} \cdot \mathbf{a})\mathbf{a} + (\mathbf{r} \cdot \mathbf{b})\mathbf{b} + (\mathbf{r} \cdot \mathbf{c})\mathbf{c}] = 0$. Since the vectors share identical magnitudes, the summation of the projection components, represented as $(\mathbf{r} \cdot \mathbf{a})\mathbf{a} + (\mathbf{r} \cdot \mathbf{b})\mathbf{b} + (\mathbf{r} \cdot \mathbf{c})\mathbf{c}$, simplifies exactly to $|\mathbf{a}|^2\mathbf{r}$. Substituting this result back into the equation yields $|\mathbf{a}|^2(3\mathbf{r} - (\mathbf{a} + \mathbf{b} + \mathbf{c})) - |\mathbf{a}|^2\mathbf{r} = 0$, which reduces to $2\mathbf{r} - (\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0$ after dividing by $|\mathbf{a}|^2$, ultimately defining $\mathbf{r}$ as $\frac{1}{2}(\mathbf{a} + \mathbf{b} + \mathbf{c})$.