Let a unit vector hatmathbfu = xhatmathbfi + yhatmathbfj + zhatmathbfk$\hat{\mathbf{u}} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}$ make angles fracpi2, fracpi3$\frac{\pi}{2}, \frac{\pi}{3}$ and frac2pi3$\frac{2pi}{3}$ with the vectors frac1sqrt2hatmathbfi + frac1sqrt2hatmathbfk$\frac{1}{\sqrt{2}}\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\hat{\mathbf{k}}$, frac1sqrt2hatmathrmj + frac1sqrt2hatmathrmk$\frac{1}{\sqrt{2}}\hat{\mathrm{j}} + \frac{1}{\sqrt{2}}\hat{\mathrm{k}}$ and frac1sqrt2hatmathbfi + frac1sqrt2hatmathrmj$\frac{1}{\sqrt{2}}\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\hat{\mathrm{j}}$ respectively. If \vec{\mathrm{v}} = \frac{1}{\sqrt{2}}\hat{\mathrm{i}} +\frac{1}{\sqrt{2}}\hat{\mathrm{j}} +\frac{1}{\sqrt{2}}\hat{\mathrm{k}}, then |\hat{\mathbf{u}} -\bar{\mathbf{v}} |^2 is equal to
A.\frac{11}{2}
B.\frac{5}{2}
C.9
D.7
Solution & Explanation
### Related Formula
vecA cdot vecB = |vecA| |vecB| cos phi$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \phi$
### Core Logic
Let the given baseline vectors be vecp_1, vecp_2, vecp_3$\vec{p}_1, \vec{p}_2, \vec{p}_3$. Note that |vecp_1| = |vecp_2| = |vecp_3| = 1$|\vec{p}_1| = |\vec{p}_2| = |\vec{p}_3| = 1$.
1. Angle with vecp_1$\vec{p}_1$ is fracpi2$\frac{\pi}{2}$:
hatu cdot vecp_1 = 0 implies fracxsqrt2 + fraczsqrt2 = 0 implies x + z = 0 quad dots (i)$\hat{u} \cdot \vec{p}_1 = 0 \implies \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \implies x + z = 0 \quad \dots (i)$
2. Angle with vecp_2$\vec{p}_2$ is fracpi3$\frac{\pi}{3}$:
hatu cdot vecp_2 = cosfracpi3 implies fracysqrt2 + fraczsqrt2 = frac12 implies y + z = frac1sqrt2 quad dots (ii)$\hat{u} \cdot \vec{p}_2 = \cos\frac{\pi}{3} \implies \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \dots (ii)$
3. Angle with vecp_3$\vec{p}_3$ is frac2pi3$\frac{2\pi}{3}$:
hatu cdot vecp_3 = cosfrac2pi3 implies fracxsqrt2 + fracysqrt2 = -frac12 implies x + y = -frac1sqrt2 quad dots (iii)$\hat{u} \cdot \vec{p}_3 = \cos\frac{2\pi}{3} \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \dots (iii)$
### Step 1: Finding Vector Coordinates
Subtracting (ii) from (iii):
(x + y) - (y + z) = -frac1sqrt2 - frac1sqrt2 implies x - z = -sqrt2 quad dots (iv)$(x + y) - (y + z) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \implies x - z = -\sqrt{2} \quad \dots (iv)$
Solving (i) and (iv):
2x = -sqrt2 implies x = -frac1sqrt2$2x = -\sqrt{2} \implies x = -\frac{1}{\sqrt{2}}$z = frac1sqrt2$z = \frac{1}{\sqrt{2}}$
From (ii): y = frac1sqrt2 - frac1sqrt2 = 0$y = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$.
So, hatu = -frac1sqrt2hati + 0hatj + frac1sqrt2hatk$\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.
### Step 2: Evaluating the Norm Difference
Given vecv = frac1sqrt2hati + frac1sqrt2hatj + frac1sqrt2hatk$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$:
hatu - vecv = left(-frac1sqrt2 - frac1sqrt2right)hati + left(0 - frac1sqrt2right)hatj + left(frac1sqrt2 - frac1sqrt2right)hatk$\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{k}$hatu - vecv = -sqrt2hati - frac1sqrt2hatj + 0hatk$\hat{u} - \vec{v} = -\sqrt{2}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$
Evaluating the squared magnitude:
|hatu - vecv|^2 = (-sqrt2)^2 + left(-frac1sqrt2right)^2 = 2 + frac12 = frac52$|\hat{u} - \vec{v}|^2 = (-\sqrt{2})^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 2 + \frac{1}{2} = \frac{5}{2}$
### Pattern Recognition
Setting up dot products systematically transforms descriptive geometric angles into solvable sets of linear equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Keywords:#Vector Dot Operations#Linear Algebra Systems#Magnitude Evaluation#JEE Main 2024 Vector Math
More Vector Algebra Previous-Year Questions — Page 2
Q69jee_main_2025_29_jan_eveningVector Products and Angles
Let hatmathbfa$\hat{\mathbf{a}}$ be a unit vector perpendicular to the vectors vecmathsfb = hatmathsfi -2hatmathsfj +3hatmathsfk$\vec{\mathsf{b}} = \hat{\mathsf{i}} -2\hat{\mathsf{j}} +3\hat{\mathsf{k}}$ and vecmathbfc = 2hatmathbfi +3hatmathbfj -hatmathbfk$\vec{\mathbf{c}} = 2\hat{\mathbf{i}} +3\hat{\mathbf{j}} -\hat{\mathbf{k}}$, and makes an angle of cos^-1left(-frac13right)$\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector hatmathrmi +hatmathrmj +hatmathrmk$\hat{\mathrm{i}} +\hat{\mathrm{j}} +\hat{\mathrm{k}}$. If hatmathbfa$\hat{\mathbf{a}}$ makes an angle of fracpi3$\frac{\pi}{3}$ with the vector hatmathrmi +alpha hatmathrmj +hatmathrmk$\hat{\mathrm{i}} +\alpha \hat{\mathrm{j}} +\hat{\mathrm{k}}$, then the value of alpha$\alpha$ is :
A.-sqrt3$-\sqrt{3}$
B.sqrt6$\sqrt{6}$
C.-sqrt6$-\sqrt{6}$
D.sqrt3$\sqrt{3}$
Solution
### Related Formula
Cross product for vector perpendicular direction alignment:
vecu = vecb times vecc$\vec{u} = \vec{b} \times \vec{c}$
Angle projection formula:
costheta = fracveca cdot vecv|veca||vecv|$\cos\theta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}||\vec{v}|}$
### Core Logic
Compute cross product of vecb$\vec{b}$ and vecc$\vec{c}$:
vecb times vecc = beginvmatrix hati & hatj & hatk \\ 1 & -2 & 3 \\ 2 & 3 & -1 endvmatrix = -7hati + 7hatj + 7hatk = -7(hati - hatj - hatk)$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = -7\hat{i} + 7\hat{j} + 7\hat{k} = -7(\hat{i} - \hat{j} - \hat{k})$
Hence, unit vector hata$\hat{a}$ matches form:
hata = pm frachati - hatj - hatksqrt3$\hat{a} = \pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$
### Step 1: Isolate Core Angle Direction
Check conditions against vector vecv = hati + hatj + hatk$\vec{v} = \hat{i} + \hat{j} + \hat{k}$:
Using hata = frachati - hatj - hatksqrt3$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$:
costheta = frac1 - 1 - 1sqrt3sqrt3 = -frac13$\cos\theta = \frac{1 - 1 - 1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3}$
This confirms the direction for hata$\hat{a}$.
### Step 2: Solve for Unknown Scalar Variable
Now compute angle with vector hati + alphahatj + hatk$\hat{i} + \alpha\hat{j} + \hat{k}$ for theta = fracpi3$\theta = \frac{\pi}{3}$:
cosfracpi3 = frac1sqrt3 cdot frac1 - alpha - 1sqrt2 + alpha^2$\cos\frac{\pi}{3} = \frac{1}{\sqrt{3}} \cdot \frac{1 - \alpha - 1}{\sqrt{2 + \alpha^2}}$frac12 = frac-alphasqrt3sqrtalpha^2 + 2$\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$
Since left hand side is positive, alpha$\alpha$ must be strictly negative. Squaring both sides:
frac14 = fracalpha^23(alpha^2 + 2) implies 3alpha^2 + 6 = 4alpha^2 implies alpha^2 = 6$\frac{1}{4} = \frac{\alpha^2}{3(\alpha^2 + 2)} \implies 3\alpha^2 + 6 = 4\alpha^2 \implies \alpha^2 = 6$
Since alpha < 0$\alpha < 0$, alpha = -sqrt6$\alpha = -\sqrt{6}$.
### Pattern Recognition
Keep strict track of signs when dealing with algebra containing square roots. Checking value constraints early on allows you to drop phantom positive/negative branches seamlessly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Q73jee_main_2025_28_jan_morningVector Dot and Cross Products
Let veca = hati +hatj +hatk,quad vecb = 2hati +2hatj +hatk$\vec{a} = \hat{i} +\hat{j} +\hat{k},\quad \vec{b} = 2\hat{i} +2\hat{j} +\hat{k}$ and vecd = vecatimes vecb$\vec{d} = \vec{a}\times \vec{b}$. If vecc$\vec{c}$ is a vector such that veca.vecc = |vecc |,$\vec{a}.\vec{c} = |\vec{c} |,$|vecc -2veca|^2 = 8$|\vec{c} -2\vec{a}|^2 = 8$ and the angle between vecd$\vec{d}$ and vecc$vec{c}$ is fracpi4,$\frac{\pi}{4},$ then |10 - 3vecb.vecc| + |vecdtimes vecc|^2$|10 - 3\vec{b}.\vec{c}| + |vec{d}\times vec{c}|^2$ is equal to ....
Q75jee_main_2025_04_april_eveningProperties of Vectors in Triangles
Let the three sides of a triangle ABC be given by the vectors 2hatmathbfi - hatmathbfj + hatmathbfk$2\hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}$ , hatmathbfi - 3hatmathbfj - 5hatmathbfk$\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$ and 3hatmathbfi - mathbf4hatmathbfj - mathbf4hatmathbfk$3\hat{\mathbf{i}} - \mathbf{4}\hat{\mathbf{j}} - \mathbf{4}\hat{\mathbf{k}}$ . Let G be the centroid of the triangle ABC. Then 6leftleft|overlineAGright|^2 + left|overlineBGright|^2 + left|overlineCGright|^2right)$6\left\left|\overline{AG}\right|^2 + \left|\overline{BG}\right|^2 + \left|\overline{CG}\right|^2\right)$ is equal to
Q64jee_main_2025_04_april_morningComponents of Vectors
Consider two vectors vecu = 3hati - hatj$\vec{u} = 3\hat{i} - \hat{j}$ and vecv = 2hati + hatj - lambda hatk$\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$, where lambda > 0$\lambda > 0$. The angle between them is given by cos^-1left(fracsqrt52sqrt7right)$\cos^{-1}\left(\frac{\sqrt{5}}{2sqrt{7}}\right)$. Let vecv = vecv_1 + vecv_2$\vec{v} = \vec{v}_1 + \vec{v}_2$, where vecv_1$\vec{v}_1$ is parallel to vecu$\vec{u}$ and vecv_2$\vec{v}_2$ is perpendicular to vecu$\vec{u}$. Then the value |vecv_1|^2 + |vecv_2|^2$|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to
A.frac232$\frac{23}{2}$
B. 14
C.frac252$\frac{25}{2}$
D. 10
Solution
### Related Formula
By orthogonal vector decomposition (Pythagorean property):
|vecv|^2 = |vecv_1|^2 + |vecv_2|^2 quad textwhen vecv_1 cdot vecv_2 = 0$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \quad \text{when } \vec{v}_1 \cdot \vec{v}_2 = 0$
### Core Logic
Compute lambda$\lambda$ using dot product formula:
costheta = fracvecu cdot vecv|vecu||vecv| implies fracsqrt52sqrt7 = frac3(2) + (-1)(1)sqrt3^2 + (-1)^2 sqrt2^2 + 1^2 + (-lambda)^2$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \implies \frac{\sqrt{5}}{2\sqrt{7}} = \frac{3(2) + (-1)(1)}{\sqrt{3^2 + (-1)^2} \sqrt{2^2 + 1^2 + (-\lambda)^2}}$fracsqrt52sqrt7 = frac5sqrt10sqrt5 + lambda^2 implies frac12sqrt7 = fracsqrt5sqrt10sqrt5 + lambda^2 = frac1sqrt2sqrt5 + lambda^2$\frac{\sqrt{5}}{2\sqrt{7}} = \frac{5}{\sqrt{10}\sqrt{5 + \lambda^2}} \implies \frac{1}{2\sqrt{7}} = \frac{\sqrt{5}}{\sqrt{10}\sqrt{5 + \lambda^2}} = \frac{1}{\sqrt{2}\sqrt{5 + \lambda^2}}$
### Step 1: Solve for lambda
Square both sides of equation:
frac128 = frac12(5 + lambda^2) implies 2(5 + lambda^2) = 28 implies 5 + lambda^2 = 14 implies lambda^2 = 9 implies lambda = 3$\frac{1}{28} = \frac{1}{2(5 + \lambda^2)} \implies 2(5 + \lambda^2) = 28 \implies 5 + \lambda^2 = 14 \implies \lambda^2 = 9 \implies \lambda = 3$
Since vecv = 2hati + hatj - 3hatk$\vec{v} = 2\hat{i} + \hat{j} - 3\hat{k}$.
### Step 2: Apply Identity
Since components are orthogonal, direct magnitude squared holds:
|vecv_1|^2 + |vecv_2|^2 = |vecv|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$
### Pattern Recognition
Do not waste time explicitly projecting components vecv_1$\vec{v}_1$ and vecv_2$\vec{v}_2$ if only the sum of their squared magnitudes is requested. The scalar length matches the total vector length invariant under any orthogonal basis change.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
More Vector Algebra Questions — jee_main_2024_29_january_evening
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