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The foot of the perpendicular from the p...

The foot of the perpendicular from the point (1,2,3) on the line `vecr=(6hati+7hatj+7hatk)+lamda(3hati+2hatj-2hatk)` has the coordinates

A

(5,8,15)

B

(8,5,15)

C

(3,5,9)

D

(3,5,-9)

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To find the foot of the perpendicular from the point \( P(1, 2, 3) \) to the line given by the vector equation \( \vec{r} = 6\hat{i} + 7\hat{j} + 7\hat{k} + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k}) \), we will follow these steps: ### Step 1: Determine the coordinates of a point on the line The line can be expressed in terms of \( \lambda \): \[ \vec{Q} = (6 + 3\lambda, 7 + 2\lambda, 7 - 2\lambda) \] This gives us the coordinates of point \( Q \) on the line in terms of \( \lambda \). ### Step 2: Find the vector \( \vec{PQ} \) The vector \( \vec{PQ} \) from point \( P(1, 2, 3) \) to point \( Q \) is given by: \[ \vec{PQ} = Q - P = (6 + 3\lambda - 1, 7 + 2\lambda - 2, 7 - 2\lambda - 3) \] This simplifies to: \[ \vec{PQ} = (5 + 3\lambda, 5 + 2\lambda, 4 - 2\lambda) \] ### Step 3: Find the direction ratios of the line The direction ratios of the line are given by the coefficients of \( \lambda \) in the line equation: \[ \vec{B} = (3, 2, -2) \] ### Step 4: Set up the dot product condition Since \( \vec{PQ} \) is perpendicular to the line, their dot product must equal zero: \[ \vec{PQ} \cdot \vec{B} = 0 \] Calculating the dot product: \[ (5 + 3\lambda) \cdot 3 + (5 + 2\lambda) \cdot 2 + (4 - 2\lambda) \cdot (-2) = 0 \] ### Step 5: Expand and simplify the equation Expanding the dot product: \[ 15 + 9\lambda + 10 + 4\lambda - 8 + 4\lambda = 0 \] Combining like terms: \[ (9\lambda + 4\lambda + 4\lambda) + (15 + 10 - 8) = 0 \] This simplifies to: \[ 17\lambda + 17 = 0 \] ### Step 6: Solve for \( \lambda \) Solving for \( \lambda \): \[ 17\lambda = -17 \implies \lambda = -1 \] ### Step 7: Substitute \( \lambda \) back to find coordinates of \( Q \) Now substitute \( \lambda = -1 \) back into the coordinates of \( Q \): \[ Q = (6 + 3(-1), 7 + 2(-1), 7 - 2(-1)) \] This simplifies to: \[ Q = (6 - 3, 7 - 2, 7 + 2) = (3, 5, 9) \] ### Conclusion The coordinates of the foot of the perpendicular from the point \( (1, 2, 3) \) to the line are \( (3, 5, 9) \). ---

To find the foot of the perpendicular from the point \( P(1, 2, 3) \) to the line given by the vector equation \( \vec{r} = 6\hat{i} + 7\hat{j} + 7\hat{k} + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k}) \), we will follow these steps: ### Step 1: Determine the coordinates of a point on the line The line can be expressed in terms of \( \lambda \): \[ \vec{Q} = (6 + 3\lambda, 7 + 2\lambda, 7 - 2\lambda) \] This gives us the coordinates of point \( Q \) on the line in terms of \( \lambda \). ...
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Find the perpendicular distance from the point (2hati-hatj+4hatk) to the plane vecr.(3hati-4hatj+12hatk) = 1 .

Find the vector equation of line passing through the point (-1, 2, 1) and parallel to the line vecr=2hati+3hatj-hatk+lamda(hati-2hatj+hatk) . Also , find the distance between these lines .

Angle between the line vecr=(2hati-hatj+hatk)+lamda(-hati+hatj+hatk) and the plane vecr.(3hati+2hatj-hatk)=4 is

Find the points of intersection of the line vecr = (2hati - hatj + 2hatk) + lambda(3hati + 4hatj + 2hatk) and the plane vecr.(hati - hatj + hatk) = 5

Find the point where line which passes through point (1, 2, 3) and is parallel to line vecr=hati-hatj+2hatk+lamda(hati-2hatj+3hatk) meets the xy -plane.

Find the distance of the point (3,8,2) from the line vecr=hati + 3 hatj + 2hatk + lamda ( 2 hati + 4 hatj + 3hatk) measured parallel to the plane vec r . (3 hati + 2 hatj - 2hatk) + 15=0.

OBJECTIVE RD SHARMA ENGLISH-PLANE AND STRAIGHT LINE IN SPACE -Chapter Test
  1. The foot of the perpendicular from the point (1,2,3) on the line vecr=...

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  2. The length of the perpendicular from the origin to the plane passing t...

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  3. The value of lamda for which the lines (x-1)/1=(y-2)/(lamda)=(z+1)/(-1...

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  4. The angle between the lines (x+4)/(1) = (y-3)/(2) = (z+2)/(3) and (x)/...

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  5. The direction cosines of the line 6x-2=3y+1=2z-2 are

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  6. A line passes through two points A(2,-3,-1) and B(8,-1,2). The coordin...

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  7. The position vector of a point at a distance of 3sqrt(11) units from h...

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  8. The line joining the points 6veca-4vecb+4vecc, -4vecc and the line joi...

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  9. The image (or reflection) of the point (1,2-1) in the plane vecr.(3hat...

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  10. The equation of the plane through the line of intersection of the plan...

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  11. Angle between the line vecr=(2hati-hatj+hatk)+lamda(-hati+hatj+hatk) a...

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  12. The line through hati+3hatj+2hatkandbot"to the line "vecr=(hati+2hatj-...

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  13. The distance of the point having position vector -hat(i) + 2hat(j) + 6...

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  14. The position vector of the point in which the line joining the points ...

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  15. The two lines vecr=veca+veclamda(vecbxxvecc) and vecr=vecb+mu(veccxxve...

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  16. Lines vecr = veca(1) + lambda vecb and vecr = veca(2) + svecb will lie...

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  17. Equation of a line passing through (-1,2,-3) and perpendicular to the ...

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  18. Find the Vector and Cartesian equation of line passing through (1, -2,...

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  19. The distance between the planes given by vecr.(hati+2hatj-2hatk)+5=0...

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  20. Find shortest distance between the line vecr = (5hati + 7hatj + 3ha...

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  21. Find the shortest distance between the lines vecr=(hatii+2hatj+hatk)+l...

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