Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

Since $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$, they represent the orthogonal vector decomposition of $\vec{v}$, satisfying the Pythagorean relation $|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2$. To determine $|\vec{v}|^2$, use the definition of the dot product $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$, where $\vec{u} = 3\hat{i}-\hat{j}$ implies $|\vec{u}| = \sqrt{3^{2+}(-1)^2} = \sqrt{10}$ and $\vec{u} \cdot \vec{v} = (3)(2) + (-1)(1) + (0)(-\lambda) = 5$.

Substituting the given angle $\cos \theta = \frac{\sqrt{5}}{2\sqrt{7}}$ into the dot product formula gives $\frac{\sqrt{5}}{2\sqrt{7}} = \frac{5}{\sqrt{10} \sqrt{5+\lambda^2}}$. Squaring both sides yields $\frac{5}{28} = \frac{25}{10(5+\lambda^2)}$, which simplifies to $\frac{5}{28} = \frac{2.5}{5+\lambda^2}$. Cross-multiplying results in $5(5+\lambda^2) = 28(2.5)$, simplifying to $25 + 5\lambda^2 = 70$. This leads to $5\lambda^2 = 45$, or $\lambda^2 = 9$. Thus, $|\vec{v}|^2 = 2^2 + 1^2 + \lambda^2 = 5 + 9 = 14$.

The condition $\vec{a}\times \vec{c}=\vec{c}\times \vec{b}$ implies that $\vec{c}$ must be parallel to the sum $\vec{a}+\vec{b}$, as rearranging the terms leads to $(\vec{a}+\vec{b})\times \vec{c}=\vec{0}$. Calculating the sum provides $\vec{a}+\vec{b} = 5\hat{i}-6\hat{j}+4\hat{k}$, allowing us to express $\vec{c}$ as $\lambda(5\hat{i}-6\hat{j}+4\hat{k})$. Expanding the given dot product $(\vec{a}+\vec{c})\cdot(\vec{b}+\vec{c})=168$ produces the expression $\vec{a}\cdot\vec{b} + \vec{c}\cdot(\vec{a}+\vec{b}) + |\vec{c}|^2 = 168$. Given that the dot product $\vec{a}\cdot\vec{b}$ equals 14 and the squared magnitude of the parallel vector is 77, we obtain the following equation:
$77\lambda^2 + 77\lambda + 14 = 168$
This simplifies to $\lambda^2 + \lambda - 2 = 0$, which gives roots $\lambda=1$ and $\lambda=-2$. Since the squared magnitude is calculated as $|\vec{c}|^2 = 77\lambda^2$, the maximum value is 308 when $\lambda=-2$.

The projection vector $\vec{c}$ of a vector $\vec{b}$ onto another vector $\vec{a}$ is given by the formula $\vec{c} = \left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right)\vec{a}$. Given $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$, its squared magnitude $|\vec{a}|^2$ is 9 and its magnitude $|\vec{a}|$ is 3. With $\vec{b} = \lambda \hat{i} + 4 \hat{k}$, the dot product $\vec{b} \cdot \vec{a}$ results in $\lambda + 8$, leading to the expression $\vec{c} = \left(\frac{\lambda + 8}{9}\right) (\hat{i} + 2\hat{j} + 2\hat{k})$.

Substituting $\vec{c}$ into the condition $|\vec{a} + \vec{c}| = 7$ allows for simplification by factoring out $\vec{a}$, so that $\vec{a} + \vec{c} = \left(1 + \frac{\lambda + 8}{9}\right)\vec{a} = \left(\frac{\lambda + 17}{9}\right)\vec{a}$. Taking the magnitude gives $\left(\frac{\lambda + 17}{9}\right) |\vec{a}| = \left(\frac{\lambda + 17}{9}\right)(3) = \frac{\lambda + 17}{3} = 7$, which confirms $\lambda = 4$.

Using $\lambda = 4$, the vectors become $\vec{b} = 4\hat{i} + 4\hat{k}$ and $\vec{c} = \frac{4}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$. The area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is the magnitude of their cross product, $|\vec{b} \times \vec{c}|$. Factoring out the scalar $\frac{4}{3}$ yields $\frac{4}{3} |\vec{b} \times \vec{a}|$. The cross product $\vec{b} \times \vec{a}$ is $-8\hat{i} - 4\hat{j} + 8\hat{k}$, which has a magnitude of 12. Multiplying this magnitude by $\frac{4}{3}$ results in an area of 16.

Intersection points between two lines in vector form are determined by equating their respective coordinates, which reveals the parameter values necessary for convergence. Equating the components of $L_1$ and $L_2$ generates the equations $\lambda - 1 = 2\mu$ and $\lambda + 1 = 3\mu + 1$, which simplify to the unique solution $\mu = 1$ and $\lambda = 3$. Substituting these into the parametric forms reveals that both lines intersect at the coordinates $(2, 8, 4)$. Since the line $L_3$ passes through this point and follows the direction vector $\vec{a} + \vec{b} = 3\hat{i} + 9\hat{j} + 4\hat{k}$, its path is defined by the equation $\vec{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + \alpha(3\hat{i} + 9\hat{j} + 4\hat{k})$. Testing the coordinates $(8, 26, 12)$ against this equation confirms that the point lies on the line, as setting the scalar parameter $\alpha = 2$ satisfies the relation for each component simultaneously.

The ratio of the magnitudes of the sum and difference of two vectors with equal magnitude $|\vec{a}|=|\vec{b}|=k$ is determined by the angle $\theta$ between them, specifically defined by $\frac{|\vec{a}+\vec{b}|}{|\vec{a}-\vec{b}|} = \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}$. Given the condition $\frac{|\vec{a}+\vec{b}| + |\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}| - |\vec{a}-\vec{b}|} = \sqrt{2}+1$, rearranging yields $\frac{|\vec{a}+\vec{b}|}{|\vec{a}-\vec{b}|} = 1+\sqrt{2}$. Squaring this ratio results in the identity $\frac{1+\cos\theta}{1-\cos\theta} = 3+2\sqrt{2}$, which allows for the isolation of $\cos\theta = \frac{1+\sqrt{2}}{2+\sqrt{2}}$. Applying this to the final expression $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2} = 2(1+\cos\theta)$ leads to the calculation $2 + 2(\frac{1+\sqrt{2}}{2+\sqrt{2}})$, which simplifies exactly to $2+\sqrt{2}$.

To find a unit vector perpendicular to two given vectors, $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 2\hat{i} + 3\hat{j} - \hat{k}$, we first compute their cross product. This operation yields the vector $-7\hat{i} + 7\hat{j} + 7\hat{k}$, which simplifies to the direction $-\hat{i} + \hat{j} + \hat{k}$. Since a unit vector must have a magnitude of one, we normalize this direction to obtain the candidates $\hat{a} = \pm \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$. We determine the correct sign by evaluating the angle $\theta$ between $\hat{a}$ and the vector $\vec{d} = \hat{i} + \hat{j} + \hat{k}$. The dot product $\hat{a} \cdot \vec{d}$ is $\pm \frac{1}{\sqrt{3}}(-1 + 1 + 1) = \pm \frac{1}{\sqrt{3}}$. Given that $\cos(\theta) = \frac{\hat{a} \cdot \vec{d}}{|\hat{a}||\vec{d}|} = \pm \frac{1/\sqrt{3}}{\sqrt{3}} = \pm \frac{1}{3}$ and the problem requires $\cos^{-1}\left(-\frac{1}{3}\right)$, we must select the negative case, resulting in $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.

With the unit vector fixed, we utilize the condition that $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\vec{e} = \hat{i} + \alpha\hat{j} + \hat{k}$. Using the definition of the dot product, the cosine of the angle is given by $\cos\left(\frac{\pi}{3}\right) = \frac{\hat{a} \cdot \vec{e}}{|\hat{a}||\vec{e}|}$, which substitutes as follows:

$\frac{1}{2} = \frac{\frac{1}{\sqrt{3}}(1 - \alpha - 1)}{1 \cdot \sqrt{1^2 + \alpha^2 + 1^2}} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$

Squaring both sides of the equation $\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$ results in $\frac{1}{4} = \frac{\alpha^2}{3(\alpha^2 + 2)}$. Cross-multiplying yields $3(\alpha^2 + 2) = 4\alpha^2$, which simplifies to $3\alpha^2 + 6 = 4\alpha^2$, leading to $\alpha^2 = 6$. Because the original equation $\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$ requires $\alpha$ to be negative to maintain the positive value of the left side, we conclude that $\alpha = -\sqrt{6}$.

The position of point $A$ on the segment $PQ$ is determined using the section formula, which states that for a point dividing a segment internally in the ratio $r:1$, the position vector is $\overrightarrow{OA} = \frac{r\overrightarrow{OQ} + \overrightarrow{OP}}{r+1}$. Using the given coordinates $P(-1, -1, 2)$ and $Q(5, 5, 10)$, we find $\overrightarrow{OA} = \left(\frac{5r-1}{r+1}\right)\hat{i} + \left(\frac{5r-1}{r+1}\right)\hat{j} + \left(\frac{10r+2}{r+1}\right)\hat{k}$.

To solve for $r$, we calculate the components of the condition $(\overrightarrow{OQ}\cdot\overrightarrow{OA})-\frac{1}{5}\left|\overrightarrow{OP}\times \overrightarrow{OA}\right|^2=10$. The dot product $\overrightarrow{OQ}\cdot\overrightarrow{OA}$ simplifies to $\frac{1}{r+1}[5(5r-1) + 5(5r-1) + 10(10r+2)] = \frac{150r+10}{r+1}$. The cross product $\overrightarrow{OP}\times \overrightarrow{OA}$ is given by the determinant

, which evaluates to $\frac{1}{r+1}(-20r\hat{i} + 20r\hat{j})$. The squared magnitude of this cross product is $\frac{(-20r)^2 + (20r)^2}{(r+1)^2} = \frac{800r^2}{(r+1)^2}$.

Substituting these expressions into the main equation yields the relation:
$\frac{150r+10}{r+1} - \frac{1}{5} \left( \frac{800r^2}{(r+1)^2} \right) = 10$
Multiplying both sides by $(r+1)^2$ results in $(150r+10)(r+1) - 160r^2 = 10(r+1)^2$. Dividing throughout by 10 simplifies the expression to $(15r+1)(r+1) - 16r^2 = (r+1)^2$. Expanding the brackets, we get $15r^2 + 16r + 1 - 16r^2 = r^2 + 2r + 1$, which further simplifies to $-r^2 + 16r + 1 = r^2 + 2r + 1$. Rearranging these terms produces $-2r^2 + 14r = 0$, and since $r > 0$, we conclude $r = 7$.

$\;\vec{A} \cdot \vec{B}=0 \;\text{ and }\;|\vec{A}|=|\vec{B}|$ $\;\Rightarrow \;\; 4-6 n+8 p=0$ $\; \;\; 3 n-4 p=2 ....(i)$ $\;\;\text{ Also }\; 4+9 n^{2+}4=4+4+16 p^2$ $\; \;\; 3 n= ± 4 p .....(ii)$ $\; \;\; ± 4 p-4 p=2$Taking - ve sign$\;-8 p=2$ $\;p=-\;\frac{1}{4}$ $\;3 n+1=2$ $\;n=\;\frac{1}{3}$

Since vector $\vec{c}$ lies in the plane defined by $\vec{a}$ and $\vec{b}$, it can be expressed as a linear combination of these two vectors in the form $\vec{c} = x\vec{a} + y\vec{b}$ for some scalar coefficients $x$ and $y$. The condition that $\vec{c}$ is perpendicular to $\vec{b}$ implies that $\vec{c} \cdot \vec{b} = 0$, while the given condition $\vec{a} \cdot \vec{c} = 5$ provides a second constraint. Using the dot products $\vec{a} \cdot \vec{b} = 2$, $|\vec{a}|^2 = 1^2 + 2^2 + 3^2 = 14$, and $|\vec{b}|^2 = 3^2 + 1^2 + (-1)^2 = 11$, these conditions generate the following system of linear equations:
$x(\vec{a} \cdot \vec{b}) + y|\vec{b}|^2 = 0 \implies 2x + 11y = 0$
$x|\vec{a}|^2 + y(\vec{a} \cdot \vec{b}) = 5 \implies 14x + 2y = 5$

Solving the first equation for $y$ gives $y = -\frac{2}{11}x$. Substituting this into the second equation yields $14x + 2(-\frac{2}{11}x) = 5$, which simplifies to $14x - \frac{4}{11}x = 5$, or $\frac{150}{11}x = 5$. This results in $x = \frac{11}{30}$ and consequently $y = -\frac{2}{11}(\frac{11}{30}) = -\frac{1}{15}$.

With these coefficients, the vector $\vec{c}$ is determined as
$\vec{c} = \frac{11}{30}(\hat{i} + 2\hat{j} + 3\hat{k}) - \frac{1}{15}(3\hat{i} + \hat{j} - \hat{k}) = \frac{1}{6}(\hat{i} + 4\hat{j} + 7\hat{k})$
The magnitude of $\vec{c}$ is then calculated by evaluating
$|\vec{c}| = \frac{1}{6} \sqrt{1^2 + 4^2 + 7^2} = \frac{1}{6}\sqrt{66} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}}$

The dot product of two unit vectors $\hat{a}$ and $\hat{b}$ is defined as $\hat{a} \cdot \hat{b} = \cos \theta$. Given $\sin \theta = \frac{\sqrt{65}}{9}$ and the range $0 < \theta < \frac{\pi}{2}$, the value of $\cos \theta$ is $\sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{65}{81}} = \sqrt{\frac{16}{81}} = \frac{4}{9}$. Since $\hat{a}$ and $\hat{b}$ are unit vectors, the dot product of either vector with itself is 1, and the dot product of either vector with the orthogonal cross product $\hat{a} \times \hat{b}$ is 0.

Distributing the dot product $\hat{a}$ across the expression $\vec{c} = 3 \hat{a} + 6 \hat{b} + 9 ( \hat{a} \times \hat{b} )$ yields $\vec{c} \cdot \hat{a} = 3(1) + 6 \cos \theta + 0 = 3 + 6 \cos \theta$. Similarly, distributing $\hat{b}$ results in $\vec{c} \cdot \hat{b} = 3 \cos \theta + 6(1) + 0 = 3 \cos \theta + 6$. Substituting these into the required expression $9 ( \vec{c} \cdot \hat{a} ) - 3 ( \vec{c} \cdot \hat{b} )$ produces $9(3 + 6 \cos \theta) - 3(3 \cos \theta + 6)$, which simplifies to $27 + 54 \cos \theta - 9 \cos \theta - 18$, or $9 + 45 \cos \theta$. Substituting $\cos \theta = \frac{4}{9}$ leads to the final result $9 + 45 \left( \frac{4}{9} \right) = 9 + 20 = 29$.

The volume of a tetrahedron is one-third the product of its base area and the altitude from the opposite vertex, a principle that allows for the determination of the altitude $h$ from vertex $D$ to the face $ABC$. Calculating the area of the triangular base $ABC$ involves finding the magnitude of the cross product of the edge vectors $\vec{AB} = \hat{j} - 3\hat{k}$ and $\vec{AC} = \hat{i} - \hat{j} - 2\hat{k}$, which results in an area of $\frac{\sqrt{35}}{2}$. Substituting this into the volume formula with the given volume of $\frac{\sqrt{805}}{6\sqrt{2}}$ identifies the altitude $h$ as $\sqrt{\frac{23}{2}}$.

Because the altitude $DE$ is perpendicular to the base, the length of the segment $AE$ follows the Pythagorean theorem within the right triangle $ADE$, where $AE^2 = AD^2 - DE^2$. Using the given $AD = \frac{\sqrt{110}}{3}$, the square of this distance becomes $AE^2 = \frac{110}{9} - \frac{23}{2} = \frac{13}{18}$. Since point $E$ lies on the median of triangle $ABC$ from $A$, the vector $\vec{AE}$ is collinear with the median vector $\vec{AM}$.

The midpoint $M$ of side $BC$ is $\frac{3\hat{i} + 4\hat{j} - 3\hat{k}}{2}$, making the vector $\vec{AM} = M - A = \frac{1}{2}(\hat{i} - 5\hat{k})$. Normalizing this vector provides the unit direction $\frac{\hat{i} - 5\hat{k}}{\sqrt{26}}$, and multiplying this by the length $AE = \sqrt{\frac{13}{18}} = \frac{\sqrt{13}}{3\sqrt{2}}$ yields $\vec{AE} = \frac{\hat{i} - 5\hat{k}}{6}$. Adding this displacement to the position vector of $A$ gives the final coordinates of $E$ as $\frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k})$.

Any vector $\vec{a}$ can be uniquely expressed as the sum of its projection along a vector $\vec{b}$ and its projection perpendicular to that same vector. By adding the provided component along $\vec{b}$, which is $\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$, to the provided component perpendicular to $\vec{b}$, which is $\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$, we determine the original vector $\vec{a}$ directly. Combining the components leads to the expression $\vec{a} = \left(\frac{48}{11} - \frac{4}{11}\right)\hat{i} + \left(\frac{16}{11} - \frac{5}{11}\right)\hat{j} + \left(-\frac{16}{11} - \frac{17}{11}\right)\hat{k}$, which simplifies to $\vec{a} = 4\hat{i} + 1\hat{j} - 3\hat{k}$. Identifying the coefficients gives $\alpha=4$, $\beta=1$, and $\gamma=-3$. Squaring these individual components results in $4^2 + 1^2 + (-3)^2 = 16 + 1 + 9$, yielding a final value of 26.

The condition $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ implies that $(\vec{b} - \vec{c}) \times \vec{d} = 0$, meaning the vector $\vec{d}$ is parallel to $\vec{b} - \vec{c}$. Computing the difference $\vec{b} - \vec{c}$ results in $\hat{i} - 2\hat{j} + \hat{k}$, so we define $\vec{d} = \lambda(\hat{i} - 2\hat{j} + \hat{k})$ for some scalar $\lambda$. Applying the condition $\vec{a} \cdot \vec{d} = 4$ with $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ leads to the equation $\lambda(1)(1) + \lambda(2)(-2) + \lambda(1)(1) = 4$, which simplifies to $-2\lambda = 4$, establishing $\lambda = -2$. Thus, the vector $\vec{d}$ is $-2\hat{i} + 4\hat{j} - 2\hat{k}$. Evaluating the cross product $\vec{a} \times \vec{d}$ as the determinant of
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix}$
produces $-8\hat{i} + 8\hat{k}$. Squaring the magnitude of this result gives $(-8)^2 + 8^2$, which equals 128.

The centroid $G$ of a triangle $ABC$ satisfies the property that the sum of the squared distances from the centroid to the vertices is given by the relation
$|\overrightarrow{AG}|^2 + |\overrightarrow{BG}|^2 + |\overrightarrow{CG}|^2 = \frac{1}{3} \left( |\vec{AB}|^2 + |\vec{BC}|^2 + |\vec{CA}|^2 \right)$
The squared magnitudes of the three given vectors representing the sides of the triangle are $|\vec{AB}|^2 = 2^2 + (-1)^2 + 1^2 = 6$, $|\vec{BC}|^2 = 1^2 + (-3)^2 + (-5)^2 = 35$, and $|\vec{CA}|^2 = 3^2 + (-4)^2 + (-4)^2 = 41$. Summing these squared magnitudes yields $6 + 35 + 41 = 82$. Substituting this value into the expression $6 \left( |\overrightarrow{AG}|^2 + |\overrightarrow{BG}|^2 + |\overrightarrow{CG}|^2 \right)$ results in $6 \left( \frac{1}{3} \times 82 \right) = 2 \times 82 = 164$.

The fact that arc $AC$ subtends a right angle at the center $O$ means the vectors $\overrightarrow{OA}$ and $\overrightarrow{OC}$ are mutually perpendicular. Since point $B$ divides the arc in a $1:5$ ratio, the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is $15^\circ$, and the angle between $\overrightarrow{OB}$ and $\overrightarrow{OC}$ is $75^\circ$. Projecting the unit vectors, we can define $\overrightarrow{OB}$ as a linear combination of $\overrightarrow{OA}$ and $\overrightarrow{OC}$ using the relation $\overrightarrow{OB} = \cos(15^\circ)\overrightarrow{OA} + \sin(15^\circ)\overrightarrow{OC}$. Substituting the trigonometric values $\cos(15^\circ) = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin(15^\circ) = \frac{\sqrt{3}-1}{2\sqrt{2}}$, we get the equation:

$2\sqrt{2}\overrightarrow{OB} = (\sqrt{3}+1)\overrightarrow{OA} + (\sqrt{3}-1)\overrightarrow{OC}$

To express $\overrightarrow{OC}$ in the required form $\alpha\overrightarrow{OA} + \beta\overrightarrow{OB}$, we rearrange this expression by isolating the $\overrightarrow{OC}$ term. Dividing by $(\sqrt{3}-1)$, we find $\overrightarrow{OC} = \frac{2\sqrt{2}}{\sqrt{3}-1}\overrightarrow{OB} - \frac{\sqrt{3}+1}{\sqrt{3}-1}\overrightarrow{OA}$. Rationalizing the denominators, the coefficient of $\overrightarrow{OA}$ becomes $-(2+\sqrt{3})$ and the coefficient of $\overrightarrow{OB}$ becomes $\sqrt{2}(\sqrt{3}+1)$. Therefore, we have $\alpha = -(2+\sqrt{3})$ and $\beta = \sqrt{2}(\sqrt{3}+1)$. Substituting these values into the target expression $\alpha + \sqrt{2}(\sqrt{3}-1)\beta$ yields $-(2+\sqrt{3}) + \sqrt{2}(\sqrt{3}-1) \cdot \sqrt{2}(\sqrt{3}+1)$, which simplifies to $-(2+\sqrt{3}) + 2(3-1) = 4 - 2 - \sqrt{3} = 2-\sqrt{3}$.