Question:
The line passing through the points \( (a, 1, 6) \) and \( (3, 4, b) \) crosses the \( yz \)-plane at \( (0, \frac{17}{2}, -\frac{13}{2}) \), then the value of \( (3a + 4b) \) is
The line passing through the points \( (a, 1, 6) \) and \( (3, 4, b) \) crosses the \( yz \)-plane at \( (0, \frac{17}{2}, -\frac{13}{2}) \), then the value of \( (3a + 4b) \) is
Show Hint
A point on the $yz$-plane always has $x=0$. Use this to find the ratio $k$ immediately.
Updated On: Apr 30, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Use Section Formula
The point on the \( yz \)-plane has \( x = 0 \). Let this point divide the segment in ratio \( k:1 \). $0 = \frac{k(3) + 1(a)}{k+1} \implies a = -3k$.
Step 2: Solve for k using y-coordinate
$y = \frac{k(4) + 1(1)}{k+1} = \frac{17}{2} \implies 8k + 2 = 17k + 17 \implies 9k = -15 \implies k = -5/3$.
Step 3: Find a and b
$a = -3(-5/3) = 5$. $z = \frac{k(b) + 1(6)}{k+1} = -13/2 \implies \frac{(-5/3)b + 6}{-2/3} = -13/2$. $-5b + 18 = 13 \implies -5b = -5 \implies b = 1$.
Step 4: Calculation
$3a + 4b = 3(5) + 4(1) = 15 + 4 = 19$.
Final Answer:(A)
The point on the \( yz \)-plane has \( x = 0 \). Let this point divide the segment in ratio \( k:1 \). $0 = \frac{k(3) + 1(a)}{k+1} \implies a = -3k$.
Step 2: Solve for k using y-coordinate
$y = \frac{k(4) + 1(1)}{k+1} = \frac{17}{2} \implies 8k + 2 = 17k + 17 \implies 9k = -15 \implies k = -5/3$.
Step 3: Find a and b
$a = -3(-5/3) = 5$. $z = \frac{k(b) + 1(6)}{k+1} = -13/2 \implies \frac{(-5/3)b + 6}{-2/3} = -13/2$. $-5b + 18 = 13 \implies -5b = -5 \implies b = 1$.
Step 4: Calculation
$3a + 4b = 3(5) + 4(1) = 15 + 4 = 19$.
Final Answer:(A)
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