Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Vector Algebra question

2025 · 28 Jan · Shift 1 · Q50
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Vector Algebra
  5. /2025 · 28 Jan · Shift 1 · Q50

Vector Algebra question

2025 · 28 Jan · Shift 1 · Q50

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let a⃗=i^+j^+k^,b→=2i^+2j^+k^\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}a=i^+j^​+k^,b=2i^+2j^​+k^ and d→=a⃗×b→\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}d=a×b. If c→\overrightarrow{\mathrm{c}}c is a vector such that a⃗⋅c→=∣c→∣\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|a⋅c=∣c∣, ∣c→−2a⃗∣2=8|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8∣c−2a∣2=8 and the angle between d→\overrightarrow{\mathrm{d}}d and c→\overrightarrow{\mathrm{c}}c is π4\frac{\pi}{4}4π​, then ∣10−3 b→⋅c→∣+∣d→×c→∣2|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2∣10−3 b⋅c∣+∣d×c∣2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6

  1. Given vectors

a⃗=(1,1,1),b⃗=(2,2,1)\vec a=(1,1,1),\qquad \vec b=(2,2,1)a=(1,1,1),b=(2,2,1)

and

d⃗=a⃗×b⃗.\vec d=\vec a\times \vec b.d=a×b.

Compute d⃗\vec dd:

d⃗=∣i^j^k^111221∣=i^(1⋅1−1⋅2)−j^(1⋅1−1⋅2)+k^(1⋅2−1⋅2)\vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1&1&1\\ 2&2&1 \end{vmatrix} =\hat i(1\cdot 1-1\cdot 2)-\hat j(1\cdot 1-1\cdot 2)+\hat k(1\cdot 2-1\cdot 2)d=​i^12​j^​12​k^11​​=i^(1⋅1−1⋅2)−j^​(1⋅1−1⋅2)+k^(1⋅2−1⋅2) d⃗=−i^+j^=(−1,1,0).\vec d=-\hat i+\hat j=( -1,1,0).d=−i^+j^​=(−1,1,0).

Also,

∣d⃗∣=(−1)2+12=2.|\vec d|=\sqrt{(-1)^2+1^2}=\sqrt2.∣d∣=(−1)2+12​=2​.

  1. Use the condition a⃗⋅c⃗=∣c⃗∣\vec a\cdot \vec c=|\vec c|a⋅c=∣c∣

Let

c⃗=(x,y,z),∣c⃗∣=r.\vec c=(x,y,z), \qquad |\vec c|=r.c=(x,y,z),∣c∣=r.

Then

x+y+z=r.\tag{1}

Now square both sides:

(x+y+z)2=x2+y2+z2.(x+y+z)^2=x^2+y^2+z^2.(x+y+z)2=x2+y2+z2.

Expanding,

x2+y2+z2+2(xy+yz+zx)=x2+y2+z2x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2x2+y2+z2+2(xy+yz+zx)=x2+y2+z2

so

xy+yz+zx=0.\tag{2}

  1. Use ∣c⃗−2a⃗∣2=8|\vec c-2\vec a|^2=8∣c−2a∣2=8

Since 2a⃗=(2,2,2)2\vec a=(2,2,2)2a=(2,2,2),

∣c⃗−2a⃗∣2=∣c⃗∣2+∣2a⃗∣2−4a⃗⋅c⃗.|\vec c-2\vec a|^2=|\vec c|^2+|2\vec a|^2-4\vec a\cdot \vec c.∣c−2a∣2=∣c∣2+∣2a∣2−4a⋅c.

Now,

∣2a⃗∣2=4∣a⃗∣2=4⋅3=12,|2\vec a|^2=4|\vec a|^2=4\cdot 3=12,∣2a∣2=4∣a∣2=4⋅3=12,

and from the condition a⃗⋅c⃗=∣c⃗∣=r\vec a\cdot\vec c=|\vec c|=ra⋅c=∣c∣=r.

Hence,

r2+12−4r=8r^2+12-4r=8r2+12−4r=8

r2−4r+4=0r^2-4r+4=0r2−4r+4=0

(r−2)2=0  ⟹  r=2. (r-2)^2=0 \implies r=2.(r−2)2=0⟹r=2.

Therefore,

|\vec c|=2, \qquad \vec a\cdot\vec c=2.\tag{3}

So from (1),

x+y+z=2.x+y+z=2.x+y+z=2.

  1. Use the angle condition between d⃗\vec dd and c⃗\vec cc

Angle between d⃗\vec dd and c⃗\vec cc is π/4\pi/4π/4, so

d⃗⋅c⃗=∣d⃗∣ ∣c⃗∣cos⁡π4.\vec d\cdot \vec c=|\vec d|\,|\vec c|\cos\frac\pi4.d⋅c=∣d∣∣c∣cos4π​.

That gives

(−1,1,0)⋅(x,y,z)=2⋅2⋅12=2,(-1,1,0)\cdot(x,y,z)=\sqrt2\cdot 2\cdot \frac{1}{\sqrt2}=2,(−1,1,0)⋅(x,y,z)=2​⋅2⋅2​1​=2,

hence

-x+y=2.\tag{4}

Also,

=2\cdot 4\cdot \frac12=4.\tag{5}$$ --- 5. **Find** $\vec b\cdot\vec c$ We need $$\vec b\cdot\vec c=2x+2y+z.$$ From $$x+y+z=2,$$ we get $$2x+2y+z=(x+y)+(x+y+z)= (x+y)+2.$$ So we need $x+y$. From (4): $$y=x+2.$$ Then using $x+y+z=2$, $$z=2-x-y=2-x-(x+2)=-2x.$$ Now use $|\vec c|^2=4$: $$x^2+y^2+z^2=4.$$ Substitute $y=x+2,\ z=-2x$: $$x^2+(x+2)^2+(-2x)^2=4$$ $$x^2+x^2+4x+4+4x^2=4$$ $$6x^2+4x=0$$ $$2x(3x+2)=0.$$ So $$x=0 \quad \text{or} \quad x=-\frac23.$$ Now use condition (2): $$xy+yz+zx=0.$$ For $x=0$: then $y=2, z=0$, and indeed $$xy+yz+zx=0.$$ For $x=-\frac23$: then $y=\frac43, z=\frac43$, giving $$xy+yz+zx=-\frac{8}{9}+\frac{16}{9}-\frac{8}{9}=0,$$ so both satisfy all conditions. Now compute $\vec b\cdot\vec c$ for both: - For $\vec c=(0,2,0)$: $$\vec b\cdot\vec c=2\cdot 0+2\cdot 2+0=4.$$ - For $\vec c=\left(-\frac23,\frac43,\frac43\right)$: $$\vec b\cdot\vec c=2\left(-\frac23\right)+2\left(\frac43\right)+\frac43= -\frac43+\frac83+\frac43=\frac83.$$ But let us use a cleaner relation. Since $$\vec b=2\vec a-(0,0,1),$$ this does not directly simplify enough. So evaluate the required expression for both valid vectors. --- 6. **Compute the required expression** We need $$|10-3\vec b\cdot\vec c|+|\vec d\times\vec c|^2.$$ From (5), $$|\vec d\times\vec c|^2=4.$$ So expression becomes $$|10-3\vec b\cdot\vec c|+4.$$ ### Case 1: $\vec c=(0,2,0)$ $$\vec b\cdot\vec c=4$$ so $$|10-3\cdot 4|+4=|10-12|+4=2+4=6.$$ ### Case 2: $\vec c=\left(-\frac23,\frac43,\frac43\right)$ $$\vec b\cdot\vec c=\frac83$$ so $$|10-3\cdot \frac83|+4=|10-8|+4=2+4=6.$$ Thus in either case, the value is $$\boxed{6}.$$
PreviousNext

More from Vector Algebra

  • If the components of a=αi^+βj^​+γk^ along and perpendicular to b=3i^+j^​−k^ respectively, are 1116​(3i^+j^​−k^) and 111​(−4i^−5j^​−17k^)…2025 · MCQ
  • Let A,B,C be three points in xy-plane, whose position vector are given by 3​i^+j^​,i^+3​j^​ and ai^+(1−a)j^​ respectively with respect to the origin O . If the distance of the point C…2025 · MCQ
  • Let a=i^+2j^​+k^ and b=2i^+7j^​+3k^. Let L1​:r=(−i^+2j^​+k^)+λa,λ∈R and L2​:r=(j^​+k^)+μb,μ∈R…2025 · MCQ
  • Let a=2i^−j^​+3k^, b=3i^−5j^​+k^ and c be a vector such that a×c=a×b=c×b and (a+c)⋅(b+c)=168…2025 · MCQ
  • Let a^ be a unit vector perpendicular to the vectors b=i^−2j^​+3k^ and c=2i^+3j^​−k^, and a^ makes an angle of cos−1(−31​) with the vector i^+j^​+k^…2025 · MCQ
  • Let a=−5i^+j^​−3k^,b=i^+2j^​−4k^ and c=(((a×b)×i^)×i^)×i^…2024 · MCQ
  • Let a=i^+j^​+k^,b=−i^−8j^​+2k^ and c=4i^+c2​j^​+c3​k^ be three vectors such that b×a=c×a…2024 · Numerical
  • Let a unit vector which makes an angle of 60∘ with 2i^+2j^​−k^ and an angle of 45∘ with i^−k^ be C. Then C+(−21​i^+32​1​j^​−32​​k^)…2024 · MCQ