Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

The area of a quadrilateral in three-dimensional space is determined by taking half the magnitude of the cross product of its diagonal vectors. Using the given vertices, the diagonal vectors are $\vec{AC} = 2\hat{i} + 5\hat{j} - 12\hat{k}$ and $\vec{BD} = 6\hat{i} + 11\hat{j} - 4\hat{k}$. Their cross product is calculated as:
$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & -12 \\ 6 & 11 & -4 \end{vmatrix} = 112\hat{i} - 64\hat{j} - 8\hat{k}$
Factoring out a common term, the magnitude of this vector is $8 \sqrt{14^2 + (-8)^2 + (-1)^2} = 8 \sqrt{261}$, and multiplying by one-half results in $\frac{1}{2} \times 8 \sqrt{9 \times 29} = 12 \sqrt{29}$.

The area of a triangle formed by vectors meeting at a common vertex is equal to half the magnitude of their cross product. By computing the cross product of the vectors $\overrightarrow{AB} = \hat{i} + 2 \hat{j} - 7 \hat{k}$ and $\overrightarrow{AC} = 6 \hat{i} + d \hat{j} - 2 \hat{k}$, we obtain the determinant

, which simplifies to $(7d-4) \hat{i} - 40 \hat{j} + (d-12) \hat{k}$. Given the area is $15 \sqrt{2}$, we apply the relation $\frac{1}{2} | \overrightarrow{AB} \times \overrightarrow{AC} | = 15 \sqrt{2}$, which leads to $(7d-4)^2 + (-40)^2 + (d-12)^2 = (30 \sqrt{2})^2$. Expanding this results in $50d^2 - 80d + 1760 = 1800$, or $5d^2 - 8d - 4 = 0$. Factoring this quadratic expression as $(5d+2)(d-2) = 0$ and noting the condition $d > 0$, we find that $d=2$.

With $d$ determined, the vector $\overrightarrow{AC}$ becomes $6 \hat{i} + 2 \hat{j} - 2 \hat{k}$. Utilizing the vector addition property $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}$, we calculate $\overrightarrow{BC} = (6-1) \hat{i} + (2-2) \hat{j} + (-2-(-7)) \hat{k} = 5 \hat{i} + 0 \hat{j} + 5 \hat{k}$. Comparing the squares of the lengths of all three sides, we find $|\overrightarrow{AB}|^2 = 1^2 + 2^2 + (-7)^2 = 54$, $|\overrightarrow{BC}|^2 = 5^2 + 0^2 + 5^2 = 50$, and $|\overrightarrow{AC}|^2 = 6^2 + 2^2 + (-2)^2 = 44$. Comparing these values, 54 is the square of the largest side.

The magnitude of the vector difference $|\vec{c} - \vec{a}| = 2\sqrt{2}$ relates to the magnitudes of $\vec{c}$ and $\vec{a}$ through the expansion $|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c} = 8$. Using the values $|\vec{a}|^2 = 6^2 + 1^2 + (-1)^2 = 38$ and the projection condition $\vec{a} \cdot \vec{c} = 6|\vec{c}|$, the relation transforms into the quadratic equation $|\vec{c}|^2 - 12|\vec{c}| + 30 = 0$. Solving for $|\vec{c}|$ yields $6 \pm \sqrt{6}$, and applying the constraint $|\vec{c}| \geq 6$ selects $|\vec{c}| = 6 + \sqrt{6}$.

Determining the cross product $\vec{a} \times \vec{b}$ results in the vector $\hat{i} - \hat{j} + 5\hat{k}$, which has a magnitude of $\sqrt{27}$. The magnitude of the cross product of this resulting vector with $\vec{c}$ is determined by the expression $|\vec{a} \times \vec{b}| |\vec{c}| \sin(60^\circ)$. Substituting the identified values, $\sqrt{27} (6 + \sqrt{6}) \frac{\sqrt{3}}{2}$, simplifies to $\frac{9}{2}(6 + \sqrt{6})$.

$\;{ {\text{R}}}^{\to }=(2 { {\text{Q}}}^{\to }+2 { {\text{P}}}^{\to })+(2 { {\text{Q}}}^{\to }-2 { {\text{P}}}^{\to })$ $\;{ {\text{R}}}^{\to }=4 { {\text{Q}}}^{\to }$Angle between $\vec{Q}$ and $\vec{R}$ is zeroOption (1) is correct

The area of a triangle defined by two vectors corresponds to half the magnitude of their cross product. Given the relationship $\vec{c} = \vec{a} - \vec{b}$, we determine vector $\vec{c}$ by subtracting the components of $\vec{b}$ from $\vec{a}$, resulting in $\vec{c} = (\alpha - 5)\hat{i} + \hat{j} - 2\hat{k}$. Calculating the cross product $\vec{a} \times \vec{c}$ yields the vector $-10\hat{i} + (4\alpha - 10)\hat{j} + (20 - 3\alpha)\hat{k}$.

Equating half the magnitude of this result to the given area $5 \sqrt{6}$ provides the condition $\sqrt{(-10)^2 + (4\alpha - 10)^2 + (20 - 3\alpha)^2} = 10 \sqrt{6}$. Squaring both sides of this equation gives $100 + (4\alpha - 10)^2 + (20 - 3\alpha)^2 = 600$, which simplifies to $(4\alpha - 10)^2 + (20 - 3\alpha)^2 = 500$. Expanding this leads to the quadratic equation $25\alpha^2 - 200\alpha = 0$, which simplifies to $\alpha(\alpha - 8) = 0$. Given that $\alpha$ must be a positive real number, we select $\alpha = 8$. Substituting this value back into the expression for $\vec{c}$ gives $3\hat{i} + \hat{j} - 2\hat{k}$, and calculating the squared magnitude $|\vec{c}|^2$ results in $3^2 + 1^2 + (-2)^2$, which equals $14$.

When two vectors have the same cross product with a common vector, the difference between the two vectors must be parallel to that common vector. Given the equality $\vec{r}\times \vec{a}=(\vec{b}+\vec{c})\times \vec{a}$, we can rearrange the expression to find that $\vec{r}-(\vec{b}+\vec{c})$ is a scalar multiple of $\vec{a}$. We can represent this relationship as $\vec{r}=\vec{b}+\vec{c}+\lambda\vec{a}$, where $\lambda$ is a scalar value that we need to determine using the second condition, $\vec{r}\cdot(\vec{b}-\vec{c})=0$.

Substituting our expression for $\vec{r}$ into the dot product equation leads to $(\vec{b}+\vec{c}+\lambda\vec{a})\cdot(\vec{b}-\vec{c})=0$, which simplifies to $\lambda(\vec{a}\cdot(\vec{b}-\vec{c})) + (\vec{b}+\vec{c})\cdot(\vec{b}-\vec{c})=0$. Recognizing that the term $(\vec{b}+\vec{c})\cdot(\vec{b}-\vec{c})$ is equivalent to $|\vec{b}|^{2-}|\vec{c}|^2$, we can use the vector components to find the necessary values. Summing the vectors gives $\vec{b}+\vec{c}=20\hat{i}+5\hat{j}-12\hat{k}$ and their difference is $\vec{b}-\vec{c}=-14\hat{i}+9\hat{j}-14\hat{k}$, while computing the dot products yields $\vec{a}\cdot\vec{b}=-389$ and $\vec{a}\cdot\vec{c}=204$. Solving for the scalar results in $\lambda = -\frac{67}{593}$.

With $\lambda$ determined, the expression for the vector becomes $\vec{r}=\vec{b}+\vec{c}-\frac{67}{593}\vec{a}$. Multiplying by 593 and rearranging terms gives $593\vec{r}+67\vec{a}=593(\vec{b}+\vec{c})$. Substituting this into the target expression $\frac{|593\vec{r}+67\vec{a}|^2}{593^2}$ simplifies the entire term to $|\vec{b}+\vec{c}|^2$. Calculating the squared magnitude of $20\hat{i}+5\hat{j}-12\hat{k}$ involves evaluating $20^{2+}5^{2+}(-12)^2$, which equals 569.

The triangle formed by the given position vectors is equilateral, as calculating the distance between each pair of vertices-$AB$, $BC$, and $CA$-consistently yields $\sqrt{2}$. In any equilateral triangle, the orthocenter coincides with the centroid, allowing us to locate this point by averaging the coordinates of the vertices to find $G \left( \frac{5}{3}, \frac{5}{3}, \frac{5}{3} \right)$.

To determine the perpendicular distance from this orthocenter to any side, we identify the midpoint $D$ of side $AB$, which is $D \left( \frac{3}{2}, 2, \frac{3}{2} \right)$. The squared length of the perpendicular $l_1$ is the squared distance between $G$ and $D$:
$l_1^2 = \left( \frac{5}{3} - \frac{3}{2} \right)^2 + \left( \frac{5}{3} - 2 \right)^2 + \left( \frac{5}{3} - \frac{3}{2} \right)^2 = \frac{1}{36} + \frac{1}{9} + \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$

Because the triangle is equilateral, the perpendicular distances to all three sides are equal, meaning $l_1 = l_2 = l_3$. Consequently, the sum of the squares of these distances is $3 \times \frac{1}{6}$, which results in $\frac{1}{2}$.

The vector triple product $(\vec{a} \times \vec{b}) \times \hat{i}$ is efficiently resolved using the identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$. Given $\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$, the respective dot products with $\hat{i}$ are $\vec{a} \cdot \hat{i} = -5$ and $\vec{b} \cdot \hat{i} = 1$. Substituting these values yields $-5\vec{b} - \vec{a}$, which expands as $-5(\hat{i} + 2\hat{j} - 4\hat{k}) - (-5\hat{i} + \hat{j} - 3\hat{k})$, resulting in $-11\hat{j} + 23\hat{k}$.

We proceed by applying the remaining cross products sequentially. The calculation $(-11\hat{j} + 23\hat{k}) \times \hat{i}$ yields $11\hat{k} + 23\hat{j}$ because $\hat{j} \times \hat{i} = -\hat{k}$ and $\hat{k} \times \hat{i} = \hat{j}$. Applying the final cross product with $\hat{i}$ gives $(11\hat{k} + 23\hat{j}) \times \hat{i}$, which resolves to $\vec{c} = 11\hat{j} - 23\hat{k}$ since $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{j} \times \hat{i} = -\hat{k}$. Finally, the dot product $\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$ involves computing $(0)(-1) + (11)(1) + (-23)(1)$, which equals $11 - 23 = -12$.

When dealing with cross products of the form $\vec{p} \times \vec{v} = \vec{a} \times \vec{p}$, the expression rearranges to $\vec{p} \times (\vec{v} + \vec{a}) = \vec{0}$, which implies that $\vec{p}$ is collinear with the vector $\vec{v} + \vec{a}$. By defining $\vec{p} = \vec{c} + \hat{i}$ and calculating $\vec{v} = \vec{a} + \vec{b} + \hat{i} = (2\hat{i} + 5\hat{j} - \hat{k}) + (2\hat{i} - 2\hat{j} + 2\hat{k}) + \hat{i} = 5\hat{i} + 3\hat{j} + \hat{k}$, the sum $\vec{v} + \vec{a}$ becomes $(5\hat{i} + 3\hat{j} + \hat{k}) + (2\hat{i} + 5\hat{j} - \hat{k}) = 7\hat{i} + 8\hat{j}$. Thus, $\vec{c} + \hat{i} = \lambda(7\hat{i} + 8\hat{j})$ for some scalar $\lambda$.

Determining $\lambda$ involves taking the dot product of both sides with $\vec{a}$, given $\vec{a} \cdot \vec{c} = -29$ and $\vec{a} \cdot \hat{i} = 2$. The left side becomes $\vec{a} \cdot (\vec{c} + \hat{i}) = -29 + 2 = -27$, and the right side becomes $\lambda(\vec{a} \cdot (7\hat{i} + 8\hat{j})) = \lambda(14 + 40) = 54\lambda$. Equating $54\lambda = -27$ yields $\lambda = -1/2$.

The final evaluation of $\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k})$ requires substituting $\vec{c} = \lambda(7\hat{i} + 8\hat{j}) - \hat{i}$ into the expression. This results in $\lambda(7\hat{i} + 8\hat{j}) \cdot (-2\hat{i} + \hat{j} + \hat{k}) - \hat{i} \cdot (-2\hat{i} + \hat{j} + \hat{k})$, which simplifies to $-\frac{1}{2}(-14 + 8) - (-2) = 3 + 2 = 5$.

$\;\cos \theta =\;\frac{5}{9}$ $\;\;\frac{\vec{a} \cdot \vec{b}}{a b}=\;\frac{5}{9} \;\; . . . . . . .(1)$ $\;|\vec{a}+\vec{b}|={\sqrt{2}}|\vec{a}-\vec{b}|$ $\;a^{2+}b^{2+}2 \vec{a} \cdot \vec{b}=2 a^{2+}2 b^{2-}4 \vec{a} \cdot \vec{b}$ $\;6 \vec{a} \cdot \vec{b}=a^{2+}b^2$ $\;6 \times \;\frac{5}{9} a b=a^{2+}b^2$ \;\;\frac{10}{3} a b=a^{2+}b^2 \;\; & \;\; a=n b $\;\;\frac{10}{3} n b^2=n^2 b^{2+}b^2$ $\;3 n^{2-}10 n+3=0$ $\;n=\;\frac{1}{3} \;\text{ and }\; n=3$ $\;\;\text{ integer value }\; n=3$

Two vectors form an acute angle if and only if their dot product is strictly greater than zero, as the dot product is the product of their magnitudes and the cosine of the angle between them. Because magnitudes are always non-negative, the sign of the cosine depends entirely on the sign of the dot product itself.

Calculating the dot product of $\alpha\hat{i}-2\hat{j}+2\hat{k}$ and $\alpha\hat{i}+2\alpha\hat{j}-2\hat{k}$ requires summing the products of corresponding components, specifically $\alpha(\alpha) + (-2)(2\alpha) + (2)(-2)$, which simplifies to $\alpha^2 - 4\alpha - 4$. For the angle to be acute, this expression must satisfy the inequality $\alpha^2 - 4\alpha - 4 > 0$. Completing the square by writing this as $(\alpha - 2)^2 - 4 - 4 > 0$ simplifies the condition to $(\alpha - 2)^2 > 8$. Taking the square root implies that $|\alpha - 2|$ must exceed $\sqrt{8}$, or approximately $2.82$. This creates two valid intervals for $\alpha$: $\alpha > 2 + 2\sqrt{2}$ or $\alpha < 2 - 2\sqrt{2}$. Since $2 + 2\sqrt{2}$ is approximately $4.82$, the smallest positive integer that satisfies this condition is 5.

In an isosceles triangle where two sides originating from a common vertex are equal, the angle bisector of that vertex angle coincides with the median to the base. Calculating the lengths of sides $AB$ and $AC$ reveals that both equal 5, which identifies $D$ as the midpoint of $BC$ with coordinates $(\frac{1}{2}, \frac{3}{2}, 3)$. Applying the distance formula for $l^2$ between $A(2, -3, 3)$ and $D(\frac{1}{2}, \frac{3}{2}, 3)$ yields the equation $l^2 = (\frac{3}{2})^2 + (-\frac{9}{2})^2 = \frac{9}{4} + \frac{81}{4} = \frac{45}{2}$. Solving for the required expression gives $2l^2 = 45$.

Distributing the vector dot product over the individual terms transforms the expression into $\vec{a}\cdot(\vec{c}\times \vec{b}) - \vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}$. The first term, $\vec{a}\cdot(\vec{c}\times \vec{b})$, represents the scalar triple product, which is equivalent to the dot product $(\vec{a}\times \vec{c})\cdot \vec{b}$. Since the problem defines $\vec{a}\times \vec{c} = \vec{b}$, this term simplifies directly to $\vec{b}\cdot \vec{b}$, or the squared magnitude $|\vec{b}|^2$. Calculating this for $\vec{b} = 3\hat{i}-3\hat{j}+3\hat{k}$ gives $3^2 + (-3)^2 + 3^2 = 27$. Next, the dot product $\vec{a}\cdot \vec{b}$ is computed as $(1)(3) + (2)(-3) + (1)(3) = 0$. By substituting these values and the given $\vec{a}\cdot \vec{c} = 3$ into the expanded expression, we obtain $27 - 0 - 3 = 24$.

The area of a parallelogram defined by vectors $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is calculated as the magnitude of their cross product, $|2\vec{a} \times 3\vec{b}|$. This expression simplifies to $6|\vec{a} \times \vec{b}|$, and because the area is given as 15 square units, we can determine that $|\vec{a} \times \vec{b}| = \frac{5}{2}$.

A quadrilateral with vertices $O, A, B, C$ has an area equal to half the magnitude of the cross product of its diagonals, $\overrightarrow{AC}$ and $\overrightarrow{OB}$. The diagonal $\overrightarrow{AC}$ is equivalent to $\overrightarrow{OC} - \overrightarrow{OA} = 3\vec{b} - 2\vec{a}$, and the diagonal $\overrightarrow{OB}$ is given as $6\vec{a} + 5\vec{b}$. Expanding the cross product $(3\vec{b} - 2\vec{a}) \times (6\vec{a} + 5\vec{b})$ results in $18(\vec{b} \times \vec{a}) - 10(\vec{a} \times \vec{b})$. Using the property $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$, this expression simplifies to $-18(\vec{a} \times \vec{b}) - 10(\vec{a} \times \vec{b}) = -28(\vec{a} \times \vec{b})$. Half the magnitude of this result is $\frac{1}{2}|-28(\vec{a} \times \vec{b})| = 14|\vec{a} \times \vec{b}|$, which, when using the previously established value of $\frac{5}{2}$, yields a total area of 35 square units.

The scalar triple product of three coplanar vectors must equal zero, as these vectors share a common plane and therefore occupy no volume. This means the determinant of the matrix formed by their components is zero, providing the geometric condition that links the variables $a, b,$ and $c$.

Setting up the determinant for the given vectors gives:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
By replacing the second column with the difference of the second and first columns, and the third column with the difference of the third and first columns, the determinant becomes:
$\begin{vmatrix} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{vmatrix} = 0$
Expanding this along the first row leads to $a(b-1)(c-1) - (1-a)(c-1) + (1-a)(b-1) = 0$. Given that $(b-1) = -(1-b)$ and $(c-1) = -(1-c)$, this simplifies to $a(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b) = 0$. Dividing the entire equation by the product $(1-a)(1-b)(1-c)$ yields $\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$. Since $\frac{a}{1-a}$ is equivalent to $\frac{a-1+1}{1-a} = -1 + \frac{1}{1-a}$, substituting this back into the equation gives $-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$, which results in the final sum being 1.

The vector triple product identity, defined by the relation $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$, provides the framework for this problem. First, calculate the left side of the equation by taking the cross product of $\vec{a} = -\hat{i}-\hat{j}+\hat{k}$ with the given vector $\vec{a} \times \vec{b} = \hat{i} - \hat{j}$, which results in $\hat{i} + \hat{j} + 2\hat{k}$. On the right side, substitute the dot product value of 1 and the squared magnitude of $\vec{a}$, which is 3, resulting in the expression $\vec{a} - 3\vec{b}$.

Equating the results gives the relationship $\vec{a} - 3\vec{b} = \hat{i} + \hat{j} + 2\hat{k}$. Multiplying this entire equation by two yields $2\vec{a} - 6\vec{b} = 2\hat{i} + 2\hat{j} + 4\hat{k}$. Isolating the required expression by subtracting $\vec{a}$ from both sides results in $\vec{a} - 6\vec{b} = (2\hat{i} + 2\hat{j} + 4\hat{k}) - \vec{a}$. Substituting the definition of $\vec{a}$ leads to $(2\hat{i} + 2\hat{j} + 4\hat{k}) - (-\hat{i} - \hat{j} + \hat{k})$, which simplifies to the following:
$\vec{a} - 6\vec{b} = 3\hat{i} + 3\hat{j} + 3\hat{k} = 3(\hat{i} + \hat{j} + \hat{k})$

The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is determined by the ratio $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. We calculate the dot product by summing the products of the corresponding components, which gives $\vec{a} \cdot \vec{b} = (5)(1) + (-1)(3) + (-3)(5) = 5 - 3 - 15 = -13$. The magnitude of the vector $\vec{b}$ is found by taking the square root of the sum of the squares of its components, resulting in $|\vec{b}| = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{35}$. Dividing the dot product by this magnitude yields a scalar projection of $\frac{-13}{\sqrt{35}}$, and since this result is negative, the corresponding projection vector is directed opposite to the reference vector $\vec{b}$.

Four points in three-dimensional space lie on the same plane if the scalar triple product of the vectors formed by connecting them is zero. We define the vectors relative to point $A$ as $\vec{AB} = B - A$, $\vec{AC} = C - A$, and $\vec{AD} = D - A$. Performing the subtraction gives $\vec{AB} = -4\hat{i}-3\hat{j}+(3-2\lambda)\hat{k}$, $\vec{AC} = -7\hat{i}+(\lambda-5)\hat{j}+(4-2\lambda)\hat{k}$, and $\vec{AD} = -6\hat{i}+0\hat{j}+(6-2\lambda)\hat{k}$. Setting the determinant of these three vectors to zero provides the coplanarity condition:
$\begin{vmatrix} -4 & -3 & 3-2\lambda \\ -7 & \lambda-5 & 4-2\lambda \\ -6 & 0 & 6-2\lambda \end{vmatrix} = 0$